Worked example 1: solution

A common first attempt

Every count matches. But H₂O₂ is hydrogen peroxide. The goal was to split water; this equation splits a different substance. Changing a subscript changed the chemistry.

Worked example 1: solution

Coefficients, not subscripts:

Worked example 1: solution

Worked example 1: the route

Worked example 2: solution

Step 2 · Balance one element at a time

O first, using the common multiple, 6: write 3 O₂ and 2 Fe₂O₃.

Fixing O changed the Fe count.

Worked example 2: solution

The recheck shows Fe unbalanced. Write 4 Fe:

Worked example 2: solution

Worked example 2: the route

Practice 1: answer B

A describes a subscript's job. C: coefficients are counts, never masses. D: multiplying every coefficient by Avogadro's number leaves the ratios unchanged, so coefficients count moles as well as molecules.

Worked example 3: solution

Step 2 · Balance one element at a time

C first: 1 = 1 already. H second: 4 H on the left need 2 H₂O.

Worked example 3: solution

O last: the right side holds 2 + 2 = 4 O, and each O₂ supplies 2. Write 2 O₂.

Worked example 3: solution

Worked example 3: the route

Worked example 4: solution

Step 2 · Balance one element at a time

C first: write 4 CO₂. Then H: write 5 H₂O.

Worked example 4: solution

O last: the right side holds 13 O. Each O₂ supplies 2, so write 13/2 O₂, then double every coefficient.

Worked example 4: solution

Worked example 4: the route

Practice 2: answer C

B stopped at the fraction: 1 + 21/2 + 8 + 5 = 24.5, not whole numbers. A doubled only the O₂: 1 + 21 + 8 + 5 = 35, with 42 O on the left against 21 on the right. D rounded 21/2 up to 11: 1 + 11 + 8 + 5 = 25, and 22 O ≠ 21.

Practice 3: answer D

O shows 2 on the left and 3 on the right; the common multiple 6 forces 3 O₂ and 2 Al₂O₃, and Al follows with 4. A fixed O but never rechecked Al: 2 ≠ 4. B balances every count, then skips the last step: every coefficient divides by 2. C balances too, but 3/2 is not a whole number; doubling it gives D.

Practice 4: answer C

A read the skeleton as already balanced: 1 + 1 + 1 = 3, but N is 2 ≠ 1. B dropped the unwritten 1 on N₂: 3 + 2 = 5. D doubled every coefficient without reducing: 2 × 6 = 12.

Practice 5: answer A

H sits in three formulas, so it goes last: set Na and O with matching 2s, and H₂'s coefficient absorbs what remains: (4 - 2)/2 = 1. B leaves H at 2 ≠ 3. C puts a 2 on everything: H becomes 4 ≠ 6. D balances, but every coefficient divides by 2.

Practice 6: solution

Names first, then balance. Only H and O need attention, and 2 H₂O settles both at once. The heating is done carefully because ammonium nitrate, heated carelessly, has a second decomposition that is considerably louder.

Worked example 1: solution

A common first attempt

Nothing burns. Combustion consumes O₂ as a reactant, and no O₂ appears on the left.

Worked example 1: solution

One reactant, two products. Only one pattern starts from a single substance: AB → A + B.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 2 · Match the shape

A fuel reacting with O₂ fits one pattern: combustion.

Worked example 2: solution

The pattern fixes both products before any balancing. Balancing now assigns the coefficients: C first, H second, O last.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1: answer C

A: a partner swap needs two compounds trading ions, and O₂ is a bare element. B: combination merges everything into one product; two products form here. D: decomposition starts from one reactant; two react here, and the fuel is not falling apart on its own.

Worked example 3: match the shape

Step 2 · Match the shape

Counting substances cannot separate these: both show two reactants and two products. The inventory can: the first equation carries a bare element, the second carries none.

Worked example 3: match the shape

Step 2 · Match the shape

Counting substances cannot separate these: both show two reactants and two products. The inventory can: the first equation carries a bare element, the second carries none.
The first fits A + BC → AC + B. The second fits AB + CD → AD + CB.

Worked example 3: name the types

Step 3 · Name the type and complete the products

Worked example 3: name the types

Step 3 · Name the type and complete the products

Worked example 3: name the types

Step 3 · Name the type and complete the products

Worked example 3: the route on the map

Practice 2: answer D

A: single displacement needs a bare element among the reactants, and every substance here is a compound. B: combination ends in one product; two form, and KNO₃ stays dissolved. C: decomposition starts from one reactant; two were mixed, and nothing split into simpler substances.

Practice 3: answer B

A: combustion consumes O₂ as a reactant; here O₂ is produced. C: the arrow reads left to right only; run backwards, the equation describes a different reaction. D: double displacement needs two compounds trading partners, and only one substance reacts.

Practice 4: answer D

A: bright and hot describes the energy released, not the type; combustion consumes O₂, and none appears here. B: a partner swap needs two compounds, and both reactants are bare elements. C: Cl₂'s bond does break, but the shape reads whole substances: one compound is built, and nothing splits into simpler substances.

Practice 5: answer B

A: double displacement needs two compounds and no bare element; Fe enters bare and Cu leaves bare. C: combination ends in one product, and two form here. D: decomposition starts from a single reactant splitting on its own; here iron does the pushing.

Practice 6: answer D

A took both names from the surface: a solid appeared, and two substances went in and two came out. B has the shape right, but the solid is silver metal pushed out of its compound, not a new ion pair leaving the water; precipitation belongs to double displacement. C has the driver right, but Mg enters as a bare element, and a partner swap needs two compounds. Single displacement is driven by electron transfer: a redox.

Worked example 1: solution

Read the ions

Worked example 1: solution

Carbonates are insoluble except with group 1 cations or NH₄⁺.

Worked example 1: solution

Carbonates are insoluble except with group 1 cations or NH₄⁺.

Worked example 1: the route on the chart

Worked example 2: testing the first attempt

A common first attempt

Worked example 2: testing the first attempt

A common first attempt

Worked example 2: the swap

Step 1 · Swap the partners Step 2 · Build each new formula

Ag⁺ pairs with Cl⁻; Na⁺ pairs with NO₃⁻. Both new pairs balance one-to-one.

Worked example 2: the swap

Step 1 · Swap the partners Step 2 · Build each new formula

Ag⁺ pairs with Cl⁻; Na⁺ pairs with NO₃⁻. Both new pairs balance one-to-one.

Worked example 2: states and the equation

Step 3 · Check each product against the rules

Worked example 2: states and the equation

Worked example 2: states and the equation

Worked example 2: the route on the chart

Practice 1: answer A

B is the soluble product: sodium is group 1 and nitrate is always soluble, so NaNO₃'s ions stay dissolved. C has the right pairing and the wrong formula: AgS gives 1(1+) + 1(2−) = −1, so charge balance needs two Ag⁺ per S²⁻. D skipped the sulfide line: Ag⁺ is not group 1 or NH₄⁺, so Ag₂S falls out.

Worked example 3: the swap

Step 1 · Swap the partners Step 2 · Build each new formula

Ba²⁺ pairs with OH⁻; Na⁺ pairs with Cl⁻. Ba²⁺ needs two OH⁻ to reach zero charge.

Worked example 3: the swap

Step 1 · Swap the partners Step 2 · Build each new formula

Ba²⁺ pairs with OH⁻; Na⁺ pairs with Cl⁻. Ba²⁺ needs two OH⁻ to reach zero charge.

Worked example 3: solution

Step 3 · Check each product against the rules

Most hydroxides are insoluble. The exception list is part of the rule, and Ba²⁺ is on it.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the chart

Practice 2: answer C

A trusted the always-soluble ions: NH₄⁺ and NO₃⁻ stay dissolved, but the new pair PbCl₂ is on the chloride exception list. B remembered only barium on the sulfate list; Sr²⁺ is there too. D applied the ammonium exception to the wrong pairing: (NH₄)₂CO₃ arrived dissolved, but NiCO₃ has no exception.

Practice 3: answer A

B stays dissolved: sodium is group 1, and Ca²⁺ is not on the chloride exception list. C is a reactant: it arrived dissolved, and its ions stay until a new pairing removes them. D skipped the carbonate line: Ca²⁺ is not a group 1 cation, so CaCO₃ falls out.

Practice 4: answer C

A pairs ammonium with nitrate: both always soluble, so NH₄NO₃ stays dissolved. B holds K⁺, a group 1 cation: always soluble. D is a reactant that arrived dissolved.

Worked example 1: the complete ionic equation

Step 2 · Write each (aq) compound as its ions

Each (aq) compound separates. The solid does not.

Worked example 1: the complete ionic equation

Na⁺ and NO₃⁻ appear identical on both sides. They never reacted.

Worked example 1: the complete ionic equation

Worked example 1: the net ionic equation

Step 4 · Check atoms and charge

Only the ions that build the solid remain.

Worked example 1: the net ionic equation

Worked example 1: the route on the strip

Worked example 2: the complete ionic equation

Step 2 · Write each (aq) compound as its ions

Pb(NO₃)₂ separates into one Pb²⁺ and two whole NO₃⁻. The coefficient on 2 KI carries through: 2 K⁺ and 2 I⁻.

Worked example 2: the complete ionic equation

Worked example 2: the complete ionic equation

Worked example 2: the net ionic equation

A common first attempt

The coefficient on I⁻ was dropped. Two iodides build each PbI₂.

Worked example 2: the net ionic equation

Worked example 2: the net ionic equation

Worked example 2: the route on the strip

Practice 1: answer C

A is the molecular equation; nothing has been written as ions. B is the complete ionic equation: Na⁺ and Cl⁻ still stand on both sides, uncancelled. D halves both charges: strontium is a group 2 metal, Sr²⁺, and carbonate is CO₃²⁻, so its 1(+1) + 1(−1) = 0 only looks balanced.

Worked example 3: solution

Step 2 · Write each (aq) compound as its ions

Worked example 3: solution

All four ions appear identical on both sides. Every ion is a spectator.

Worked example 3: solution

Worked example 3: the route on the strip

Practice 2: answer A

B dropped the coefficient: Ag 1 ≠ 2, and charge 1(1+) + 1(2−) = −1 against 0. C precipitated the soluble pair: K⁺ is group 1 and nitrate is always soluble. D is the complete ionic equation, K⁺ and NO₃⁻ uncancelled. E copied a one-to-one ratio into the formula: AgCO₃ sums to 1(1+) + 1(2−) = −1, not neutral.

Worked example 4: the complete ionic equation

Step 2 · Write each (aq) compound as its ions

All three (aq) compounds separate completely. Water is a covalent molecule: it stays whole.

Worked example 4: the complete ionic equation

Worked example 4: the complete ionic equation

Worked example 4: the net ionic equation

Step 4 · Check atoms and charge

Worked example 4: the net ionic equation

Step 4 · Check atoms and charge

Worked example 4: the net ionic equation

Step 4 · Check atoms and charge

Worked example 4: a weak acid instead

The weak-acid case

A weak acid is not written as ions: the water really holds HCN molecules.

Worked example 4: a weak acid instead

The weak-acid case

A weak acid is not written as ions: the water really holds HCN molecules.

Worked example 4: a weak acid instead

The weak-acid case

A weak acid is not written as ions: the water really holds HCN molecules.

Worked example 4: the route on the strip

Worked example 5: solution

Step 2 · Write each (aq) compound as its ions

The solid metal and the gas stay whole; only the two (aq) compounds separate.

Worked example 5: solution

Worked example 5: solution

Worked example 5: the route on the strip

Practice 3: answer A

B dropped the coefficient: 1(3+) + 1(1−) = 2+ against 0, and one OH⁻ cannot supply the three in Fe(OH)₃. C is the complete ionic equation: Na⁺ and Cl⁻ still stand on both sides. D is the molecular equation; nothing has been written as ions.

Practice 4: answer D

B builds the wrong product: KCl passes the solubility rules, so its ions stay dissolved and KCl(s) never forms. C is the complete ionic equation with K⁺ and Cl⁻ uncancelled. A halves both charges: barium is a group 2 metal, Ba²⁺, and carbonate is CO₃²⁻; its 1(1+) + 1(1−) = 0 only looks balanced.

Worked example 1: solution

Step 1 · Classify the solute

A metal with a nonmetal: ionic. A soluble ionic compound is a strong electrolyte.

Worked example 1: solution

The subscript counts two separate chloride ions. Each one leaves the lattice on its own.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the chart

Worked example 2: classifying

Step 1 · Classify the solute

Worked example 2: classifying

Step 1 · Classify the solute

Worked example 2: what each solution contains

Step 2 · Write what water makes Step 3 · Check the charge sum

Worked example 2: what each solution contains

Step 2 · Write what water makes Step 3 · Check the charge sum

Worked example 2: what each solution contains

Step 2 · Write what water makes Step 3 · Check the charge sum

Worked example 2: what each solution contains

Step 2 · Write what water makes Step 3 · Check the charge sum

Worked example 2: the route on the chart

Practice 1: answer B

A: solubility decides, not the anion; a soluble salt dissociates completely, polyatomic ion or not. C: the subscript counts three separate K⁺ ions; no K₃⁺ unit exists, and 1 + 1 = 2 undercounts the ions. D: an ionic compound has no molecules; only separated ions enter the water.

Worked example 3: solution

Step 1 · Classify the solute

A metal with a nonmetal: soluble ionic, a strong electrolyte. Every formula unit dissociates.

Worked example 3: solution

Worked example 3: solution

Each liter holds 0.100 mol of dissolved AlBr₃, and each mole releases 3 mol of Br⁻.

Worked example 3: solution

Each liter holds 0.100 mol of dissolved AlBr₃, and each mole releases 3 mol of Br⁻.

Worked example 3: the route on the chart

Practice 2: answer D

C skipped the ratio: 0.60 × 1 = 0.60 M assumes one sulfate per formula unit. B flipped the ratio: 0.60 × 3 = 1.8 M, a tank at 1.8 × 3 = 5.4 M sulfate. A used iron's subscript: 0.60 ÷ 2 = 0.30 M.

Worked example 4: solution

A common first attempt

Dissolving freely is not ionizing. Mixing spreads molecules through the water; only reaction with water makes ions.

Worked example 4: solution

HNO₃ is on the list: a strong electrolyte. HC₂H₃O₂ is not, and an acid off the list is weak.

Worked example 4: solution

Worked example 4: solution

Worked example 4: the route on the chart

Practice 3: answer A

B is on the strong-acid list (HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄, HClO₃): every molecule ionizes, and the bulb glows bright. C is molecular and neither acid nor base: sucrose dissolves as whole molecules, and the bulb stays dark. D shares fluorine with HF but is a soluble ionic compound: NaF → Na⁺ + F⁻ dissociates completely, a bright bulb.

Worked example 1: solution

HNO₃ · on the strong-acid list

HNO₃ is one of the seven strong acids, so it ionizes completely: a strong electrolyte.

Worked example 1: solution

HNO₃ · on the strong-acid list

HF is not one of the seven, so it is a weak acid: a weak electrolyte, partial ionization.

Worked example 1: solution

HNO₃ · on the strong-acid list

Worked example 1: solution

HNO₃ · on the strong-acid list

Worked example 1: the route on the chart

Practice 1: answer C

Watch the chlorine oxoacids: HClO₄ (perchloric) and HClO₃ (chloric) are on the strong list, but HClO (hypochlorous) is weak: it ionizes only partially (⇌). HCl and HNO₃ are strong (→).

Practice 2: answer B

A assumes every acid ionizes fully; only the seven on the list do, and HNO₂ is not one. C reads dissolves completely as ionizes completely; both dissolve fully, but only the salt separates fully. D misclassifies NaNO₂: a metal with a polyatomic anion is ionic, and a soluble ionic compound always dissociates.

Practice 3: answer A

B ranks by electronegativity; the list of seven is measured behavior, and HF is not on it. C confuses concentration with strength: both bottles hold 1 M of acid, but they deliver very different amounts of H₃O⁺. D inverts the vocabulary: weak describes the fraction ionized, not a loose grip on H⁺; the weak acid is the one that keeps its H.

Worked example 2: the products

Predict the products

The salt pairs Ca²⁺ with PO₄³⁻; charge balance builds Ca₃(PO₄)₂. The other product of every neutralization is water.

Worked example 2: the products

Predict the products

The salt pairs Ca²⁺ with PO₄³⁻; charge balance builds Ca₃(PO₄)₂. The other product of every neutralization is water.

Worked example 2: the products

Predict the products

The salt pairs Ca²⁺ with PO₄³⁻; charge balance builds Ca₃(PO₄)₂. The other product of every neutralization is water.

Worked example 2: the coefficients

Match H⁺ to OH⁻

Balancing a neutralization needs equal moles of H⁺ and OH⁻: one water forms per pair. Three H⁺ against two OH⁻ meet at six: write 2 H₃PO₄ and 3 Ca(OH)₂.

Worked example 2: the coefficients

Worked example 2: the coefficients

Worked example 2: the route on the chart

Worked example 3: the exchange

First reaction · exchange partners

Worked example 3: the exchange

First reaction · exchange partners

Worked example 3: the exchange

First reaction · exchange partners

Worked example 3: the gas appears

Second reaction · the unstable product breaks up

Worked example 3: the gas appears

Worked example 3: the gas appears

Worked example 3: the route on the chart

Practice 4: answer D

A stopped at the exchange: H₂CO₃ breaks into H₂O and CO₂ the moment it forms. B expects H₂, the gas an acid releases from an active metal; a bicarbonate trades partners instead. C reads weak as unreactive: weak describes the fraction ionized, and acid plus a carbonate or bicarbonate always gives CO₂, water, and a salt.

Guided example: solution

Step 1 · Assign the known numbers

Hydrogen's rule gives +1. Carbon has no rule of its own: call it x.

Guided example: solution

Four H atoms contribute 4(+1) = +4. Methane is neutral, so the sum is 0.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Worked example 1: solution

Step 1 · Assign the known numbers

Oxygen's rule gives −2. Sulfur has no rule of its own: call it x.

Worked example 1: solution

Two O atoms contribute 2(−2) = −4. The compound is neutral, so the sum is 0.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Assign the known numbers

Potassium is Group 1: +1. Oxygen's rule gives −2. Manganese has no rule of its own: call it x.

Worked example 2: solution

Four O atoms contribute 4(−2) = −8. The compound is neutral, so the sum is 0.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1: answer D

A left out calcium: x + 3(−2) = 0 gives +6, and +2 + 6 − 6 = +2, not 0. B gave S the sulfite ion's charge, −2; that charge belongs to the whole SO₃ group. C gave Ca Group 1's +1: +1 + x − 6 = 0 → +5, but Ca sits in Group 2. E counted the −2 once: +2 + x − 2 = 0 → 0; three O atoms contribute −6.

Practice 1: answer D

Worked example 3: testing the first attempt

A common first attempt

Worked example 3: testing the first attempt

A common first attempt

Worked example 3: solution

Step 2 · Multiply by the subscripts Step 3 · Set the sum equal to the charge

Four O atoms contribute 4(−2) = −8. The species is an ion, so the sum is its charge: −2.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 2: answer A

B set the sum to zero: 2x − 8 = 0 → +4, and 2(+4) − 8 = 0 ≠ −2. C used +2 for the 2− charge: 2x − 8 = +2 → +5. D dropped the subscript on C: x − 8 = −2 → +6, but two C atoms at +6 sum to 12 − 8 = +4.

Practice 2: answer A

Practice 3: answer B

A set the ion's sum to zero: x + 4 = 0 gives −4, and −4 + 4 = 0 ≠ +1. C gave N the ion's charge; that shortcut holds only for a monatomic ion. D flipped the sign: +3 + 3(+1) = +6, not 0. H already holds the positive side, so N balances from the negative side.

Practice 3: answer B

Practice 4: answer C

A set the sum to zero: x + 4(−2) = 0 → +8. B put the unknown on the negative side: −5 − 8 = −13, not −3. D counted the −2 once for the whole ion: x + (−2) = −3 → −1; four O atoms contribute −8.

Practice 4: answer C

Guided example: solution

Step 2 · Find the changes

Fe rises 0 → +2: iron is oxidized, losing 2 electrons. S falls 0 → −2: sulfur is reduced, gaining 2 electrons.

Guided example: solution

Fe gave its electrons to S: Fe is the reducing agent. S took them: S is the oxidizing agent.

Worked example 1: solution

Step 2 · Find the changes

Zn increases, 0 → +2: zinc is oxidized. It lost 2 electrons. Cu decreases, +2 → 0: copper is reduced. It gained 2 electrons.

Worked example 1: solution

S stays +6 and O stays −2. The sulfate group moves as one piece; no atom in it loses or gains anything. The transfer runs between Zn and Cu only.

Worked example 1: solution

Worked example 2: solution

Step 2 · Find the changes

Cu increases, 0 → +2: copper is oxidized, losing 2 electrons. Ag decreases, +1 → 0: silver is reduced, each atom gaining 1 electron. N and O do not change.

Worked example 2: solution

Copper is oxidized. That does not make it the oxidizing agent: the name describes what a reactant does to its partner. Copper gives its electrons away and oxidizes nothing. The silver in AgNO₃ takes them, so AgNO₃ does the oxidizing.

Worked example 2: solution

Worked example 2: solution

Practice 1: answer B

A named the oxidized species: losing electrons to the partner is exactly what makes Ni the reducing agent. C followed the partner-swapping: Cl is −1 on both sides, a change of (−1) − (−1) = 0, so chloride transferred nothing. D swapped both names: CuCl₂ takes nickel's electrons, and the taker does the oxidizing.

Practice 1: answer B

Worked example 3: solution

Step 1 · Assign oxidation numbers

K, a group 1A metal, is +1; the iodide ion is −1; O is −2; nitrate's sum, x + 3(−2) = −1, gives N +5; Pb balances each formula at +2.

Worked example 3: solution

Step 1 · Assign oxidation numbers

Every number is the same on both sides. Nothing is oxidized and nothing is reduced: no electrons moved. This reaction is not redox, and it has no oxidizing or reducing agent.

Worked example 3: solution

Step 1 · Assign oxidation numbers

Worked example 4: solution

The oxidation half-reaction

Zn rises 0 → +2: it lost 2 electrons. Lost electrons are written as a product.

Worked example 4: solution

Cu falls +2 → 0: it gained 2 electrons. Gained electrons are written as a reactant.

Worked example 4: solution

Practice 2: answer D

Fe comes from the sums: x + 2(−1) = 0 gives +2; x + 3(−1) = 0 gives +3. A swapped the names: Cl₂ takes iron's electrons, and the taker is the oxidizing agent. B added lost to gained, 2 + 2 = 4; they are the same 2 electrons, changing hands once. C counted one Fe atom, 3 − 2 = 1 e⁻, but the equation oxidizes 2 Fe.

Practice 2: answer D

Practice 3: answer E

A is the other half: Ag⁺ gains the electrons, so it belongs to AgNO₃, the oxidizing agent. B put lost electrons on the gain side: left 0 + 3(1−) = −3, right 3+. C copied one silver's single electron: right (3+) + 1(1−) = +2, not 0. D balances, (3+) + 3(1−) = 0, but runs backward: Al³⁺ is made here, not used.

Practice 3: answer E

Practice 4: answer C

A named the oxidized species: KBr gives its electrons away, which makes it the reducing agent. B followed the partner-swapping: K is +1 in KBr and +1 in KCl, a change of 1 − 1 = 0. D swapped both names: Cl₂ takes bromide's electrons, and the taker does the oxidizing. An oxidizing agent need not contain a metal.

Practice 4: answer C

Worked example 1: solution

Step 2 · Compare positions

Mg stands above hydrogen. The reaction runs: magnesium hands its electrons to H⁺.

Worked example 1: solution

Metal + acid → salt + H₂. Magnesium forms a 2+ ion and chloride is 1−, so the salt is MgCl₂.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the series

Worked example 2: solution

Step 2 · Compare positions

Al stands above Ni. The reaction runs: aluminum displaces the nickel ion from solution.

Worked example 2: solution

Aluminum forms a 3+ ion; nitrate is 1−. Three nitrates balance one aluminum, and nickel leaves as the solid metal.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the series

Practice 1: answer A

B reverses the ranking: zinc stands in the steam tier and lead in the acid tier below it. C has the displacement right and the salt wrong: zinc forms a 2+ ion and nitrate is 1−, so the salt is Zn(NO₃)₂, (2+) + 2(1−) = 0. D runs the series upward: lead, the lower metal, never displaces Zn²⁺.

Worked example 3: solution

A common first attempt

Worked example 3: solution

Cu sits far below Li. Copper keeps its electrons, and Li⁺ stays dissolved.

Worked example 3: solution

Worked example 3: the route on the series

Practice 2: answer B

A treats balance as permission: Ni 1 = 1, Fe 1 = 1, S 1 = 1, O 4 = 4, yet a balanced equation can describe a reaction that never happens. C compared each metal with hydrogen; the partner here is Fe²⁺, which sits above Ni. D: speed never moves a metal up the series; the ranking decides whether electrons flow at all.

Practice 3: solutions

Practice 3: solutions

Worked example 1: solution

One conversion factor is needed.

Step 2 · Build the ratio, wanted over given

Two orientations exist. Only one cancels the given unit:

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Practice 1: answer B

A assumed 1:1: 7.00 × 1 = 7.00. C inverted the ratio: 7.00 × (1/2) = 3.50. D answered with the coefficient, 2.

Practice 2: answer C

A assumed 1:1: 4.8 × 1 = 4.8. B inverted the ratio: 4.8 × (3/2) = 7.2. D doubled the given: 4.8 × 2 = 9.6.

Worked example 2: solution

One conversion factor is needed.

Step 2 · Build the ratio, wanted over given

The ratio, wanted on top: 2 mol KClO₃ over 3 mol O₂, so mol O₂ cancels.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 3: answer C

A assumed a 1 : 1 ratio: 26.0 × 1 = 26.0. B multiplied by 2 without dividing by 8: 26.0 × 2 = 52.0. D inverted the ratio: 26.0 × (8/2) = 104.

Practice 4: answer D

Practice 4: answer D

A stopped at the CO₂ alone: 1.50. B assumed 1:1 for each product: 2.40 + 2.40 = 4.80. C inverted both ratios: 2.40 × (8/5) + 2.40 × (8/6) = 3.84 + 3.20 = 7.04.

Worked example 1: solution

Two conversion factors are needed.

Step 1 · Grams → moles

CH₄'s own molar mass converts the given mass to moles:

Worked example 1: solution

The mole ratio, written CO₂ over CH₄ (1 : 1), crosses substances:

Worked example 1: solution

Worked example 1: the route on the map

Worked example 1: one more factor

Step 3 · Moles → grams

CO₂'s molar mass converts the moles out to mass. The same chain, one factor longer:

Worked example 1: one more factor

Step 3 · Moles → grams

CO₂'s molar mass converts the moles out to mass. The same chain, one factor longer:

Worked example 1: the grams route on the map

Worked example 2: solution

Three conversion factors are needed.

Step 1 · Grams → moles

Glucose's own molar mass converts the given mass to moles. Its unit cancels the given unit:

Worked example 2: solution

The mole ratio is the only factor that crosses substances. Two orientations exist. Only one cancels mol glucose:

Worked example 2: solution

Convert out with the target's molar mass, in one continuous setup. Carry all digits and round once at the end:

Worked example 2: solution

Worked example 2: the route on the map

Worked example 3: solution

Three conversion factors are needed.

Step 1 · Grams → moles

NaN₃'s molar mass, 65.02 g per mole, converts 50.0 g to 0.769 mol NaN₃.

Worked example 3: solution

The mole ratio, written N₂ over NaN₃ (3 : 2), crosses substances and turns 0.769 mol NaN₃ into 1.1535 mol N₂, unrounded.

Worked example 3: solution

Avogadro's number converts out, with mol N₂ on the bottom so it cancels. Carry all digits and round once at the end:

Worked example 3: solution

Run backward, the chain starts at a count: molecules B → mol B → mol A → g A, with Avogadro's number flipped so molecules cancel.

Worked example 3: the route on the map

Practice 1: answer C

A skipped the mole ratio: 25.0/44.09 × 18.02 = 10.2. B stopped at moles: 25.0/44.09 × 4 = 2.27 mol H₂O, one factor short of grams. D inverted the ratio: 25.0/44.09 × (1/4) × 18.02 = 2.55.

Practice 2: answer B

A stopped at moles: 35.0/26.98 × (2/4) = 0.649 mol Al₂O₃, one factor short of grams. C skipped the mole ratio: 35.0/26.98 × 101.96 = 132. D inverted the ratio: 35.0/26.98 × (4/2) × 101.96 = 265.

Practice 3: answer D

A skipped the mole ratio: 2.20/44.01 × 84.01 = 4.20. B inverted the ratio: 2.20/44.01 × (1/2) × 84.01 = 2.10. C stopped at moles: 2.20/44.01 × 2 = 0.100 mol NaHCO₃, one factor short of grams.

Worked example 4: solution

Three conversion factors are needed.

Step 1 · Grams → moles

The molar mass, 28.09 g per mole, converts the given mass of Si to moles.

Worked example 4: solution

The mole ratio, written Cr over Si (4 : 3), crosses substances; mol Si cancels. Only the substances changed; the route is the same.

Worked example 4: solution

Convert out with Cr's molar mass. The complete chain:

Label every quantity with its substance. A wrong setup shows up as a unit that will not cancel.

Worked example 4: solution

Worked example 4: the route on the map

Worked example 1: which reactant limits

Step 1 · Convert both reactants to moles

Both amounts are already in moles, so Step 1 is complete.

Step 2 · Divide each by its coefficient

Worked example 1: which reactant limits

Worked example 1: which reactant limits

Worked example 1: how much product

Step 4 · Compute product from the limiting reactant only

Worked example 1: how much product

Step 4 · Compute product from the limiting reactant only

Worked example 1: the route on the map

Worked example 2: which reactant limits

Four conversion factors are needed: one molar mass for each reactant, then the mole ratio, then the target's molar mass.

Step 1 · Convert both reactants to moles

Worked example 2: which reactant limits

Worked example 2: which reactant limits

Worked example 2: how much product

Step 4 · Compute product from the limiting reactant only

Worked example 2: how much product

Step 4 · Compute product from the limiting reactant only

Worked example 2: the route on the map

Worked example 3: the NH₃ each reactant can make

Three conversion factors per reactant: its molar mass, the mole ratio, and the molar mass of NH₃.

Step 1 · Grams to moles, for each reactant

Worked example 3: the NH₃ each reactant can make

Worked example 3: the NH₃ each reactant can make

Worked example 3: grams of NH₃, then compare

Step 3 · Moles to grams of product

Worked example 3: grams of NH₃, then compare

Step 3 · Moles to grams of product

Worked example 3: the route on the map

Worked example 3: both paths agree

Worked example 3: both paths agree

Worked example 4: solution

Consumed by the limiting reactant

The mole ratio converts the Al that reacts into the Fe₂O₃ it consumes:

Worked example 4: solution

Consumed by the limiting reactant

The mole ratio converts the Al that reacts into the Fe₂O₃ it consumes:

Worked example 4: solution

Consumed by the limiting reactant

The mole ratio converts the Al that reacts into the Fe₂O₃ it consumes:

Worked example 4: the route on the map

Practice 1: answer A

B stopped at moles: 30.0/70.90 × (2/3) = 0.282 mol AlCl₃, one factor short of grams. C skipped the mole ratio: 30.0/70.90 × 133.33 = 56.4. D inverted the ratio: 30.0/70.90 × (3/2) × 133.33 = 84.6.

Practice 2: answer B

A took the smaller mass, H₂, as limiting: 3.00/2.016 × (1/2) × 32.04 = 23.8. C skipped the grams-to-moles step and treated grams of CO as moles: 14.0 × 32.04 = 449. D stopped at moles: 14.0/28.01 × (1/1) = 0.500 mol CH₃OH, one factor short of grams.

Practice 3: answer C

Practice 3: answer C

A kept the larger amount, from CO₂, the excess (also the smaller mass and fewer moles): 53.0. B stopped at moles: 0.900 × (1/2) = 0.450. D inverted the ratio: 0.900 × 2 × 105.99 = 191. E skipped the ratio: 0.900 × 105.99 = 95.4.

Practice 4: answer A

B compared raw moles, called P₄ limiting, and built product from it: 0.50 × (4/1) = 2.0. C carried the 2.4 mol of Cl₂ across as 1:1: 2.4 × 1 = 2.4. D inverted the ratio: 2.4 × (6/4) = 3.6.

Practice 5: answer D

Practice 5: answer D

A stopped at the Mg consumed, 12.82 g. B inverted the ratio: 20.00 − 50.00/189.67 × (1/2) × 24.31 = 16.80. C skipped it: 20.00 − 50.00/189.67 × 24.31 = 13.59.