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 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

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: write 4 CO₂. Then H: write 5 H₂O.

Worked example 3 — solution

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

Worked example 3 — solution

Practice 2 — answer: C

C first (2 CO₂), H next (2 H₂O), O last: the right side holds 2 × 2 + 2 = 6, so 3 O₂. A omitted the unwritten 1: 3 + 2 + 2 = 7. B doubled every coefficient without reducing: 2 × 8 = 16. D doubled only the products: 1 + 3 + 4 + 4 = 12, and the O count no longer balances.

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 2 — solution

Step 2 · Build the ratio, wanted over given

The wanted substance is O₂, so the ratio is written O₂ over C₂H₂.

Worked example 2 — solution

Step 2 · Build the ratio, wanted over given

The wanted substance is O₂, so the ratio is written O₂ over C₂H₂.
Step 3 · Multiply and cancel

Worked example 2 — solution

Step 2 · Build the ratio, wanted over given

The wanted substance is O₂, so the ratio is written O₂ over C₂H₂.
Step 3 · Multiply and cancel

Practice 1 — answer: B

A assumed 1:1: 7.0 × 1 = 7.0. C inverted the ratio: 7.0 × (1/2) = 3.5. 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 3 — 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 3 — solution

Worked example 3 — solution

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.

Worked example 1 — 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 1 — solution

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

Worked example 1 — solution

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

Worked example 1 — solution

Worked example 2 — 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 2 — 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 2 — 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 2 — solution

Practice 1 — answer: C

A skipped the mole ratio: 25.0/44.09 × 18.02 = 10.2. B applied the ratio to grams: 25.0 × 4 = 100. D inverted the ratio: 25.0/44.09 × (1/4) × 18.02 = 2.55.

Practice 2 — answer: B

A applied the ratio to grams: 35.0 × (2/4) = 17.5. 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.

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 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

Practice 1 — answer: C

A applied the ratio to grams: 25.0 × (1/2) = 12.5. B built product from the excess Mg: 10.0/24.31 × 95.21 = 39.2. D skipped the mole ratio: 25.0/36.46 × 95.21 = 65.3.

Practice 2 — answer: A

B built product from the excess propane: 11.0/44.09 × 3 × 44.01 = 32.9. C skipped the mole ratio: 1.00 × 44.01 = 44.0. D applied the ratio to grams: 32.0 × (3/5) = 19.2.

Worked example 1 — solution

Two conversion factors are needed.

Step 1 · Volume → moles

250.0 mL is 0.2500 L. The label's molarity converts the liters to moles of HCl:

Worked example 1 — solution

The mole ratio, written H₂ over HCl (1 : 2), crosses substances; mol HCl cancels:

Worked example 1 — solution

Worked example 2 — solution

Three conversion factors are needed.

Step 1 · Volume → moles

150.0 mL is 0.1500 L. The molarity, 0.300 mol per liter, converts the liters to 0.0450 mol KI.

Worked example 2 — solution

mol KI cannot cancel mol PbI₂. The equation makes one PbI₂ from two KI; the mole ratio must cross first.

Worked example 2 — solution

The ratio, written PbI₂ over KI (1 : 2), crosses substances; the molar mass then converts out:

Worked example 2 — solution

Practice 1 — answer: A

B skipped the mole ratio: 0.0600 × 106.87 = 6.41. C inverted the ratio: 0.0600 × 3 × 106.87 = 19.2. D fed milliliters to the molarity: 300.0 × 0.200 = 60.0, a mole count 1000 times too large, and the chain ends at 2140 g.

Worked example 3 — solution

Three conversion factors are needed. Molarity works at both ends: the base's converts volume in to moles, the acid's converts moles out to volume.

Step 1 · Volume → moles

175.0 mL is 0.1750 L. The base's molarity converts the liters to moles of NaOH:

Worked example 3 — solution

The mole ratio, written H₂SO₄ over NaOH (1 : 2), gives 0.0210 mol H₂SO₄.

Worked example 3 — solution

The acid's molarity converts moles out to volume:

Worked example 3 — solution

Practice 2 — answer: D

A flipped the acid's molarity factor: 0.0750 × 0.400 = 0.0300, and the units, mol²/L, are not a volume. B inverted the mole ratio: 0.0375 × 1/2 ÷ 0.400 = 0.0469 L, or 46.9 mL. C skipped the mole ratio: 0.0375 ÷ 0.400 = 0.0938 L, or 93.8 mL.

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 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

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

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.

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 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

Practice 1 — answer: B

A is the soluble product: potassium is group 1 and nitrate is always soluble, so KNO₃'s ions stay dissolved. C is a reactant: Ba(NO₃)₂ arrived dissolved, and a dissolved compound does not leave solution when another is added. D skipped the check on BaSO₄ — barium is on the sulfate exception list.

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

Practice 2 — answer: C

A: chlorides are soluble except with Ag⁺, Pb²⁺, Hg₂²⁺, and Mg²⁺ is not among them, so MgCl₂ stays dissolved. B: sodium is group 1, and group 1 compounds are always soluble. D is a reactant: MgSO₄ arrived dissolved, and no new pairing removed its ions.

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 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

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

Practice 2 — answer: A

B dropped the coefficient: charge 1(+3) + 1(−1) = +2 against 0, and one OH cannot supply the three in Al(OH)₃. C is the complete ionic equation with Na⁺ and NO₃⁻ uncancelled. D is the molecular equation; no compound has been written as its ions.

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 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

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

A common first attempt

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

Worked example 3 — solution

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

Worked example 3 — solution

Worked example 3 — solution

Practice 2 — answer: C

A: the list is HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄; HF is not on it, and an acid off the list is weak. B: a small fraction does ionize, so the solution conducts weakly; a nonelectrolyte would not conduct at all. D: H⁺ and H₃O⁺ are two names for the same ion, so 1 + 1 + 1 = 3 counts it twice; each ionized HF gives 1 + 1 = 2 ions.

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 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 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

Practice 1 — answer: B

A set the sum to zero: x + 4(−2) = 0 → +8, a neutral species' answer. C counted the −2 once for the whole ion: x + (−2) = −2 → 0; four O atoms contribute −8. D put the unknown on the negative side: −6 − 8 = −14, not −2.

Practice 2 — 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.

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.

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

Practice 2 — 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.

Worked example 1 — solution

Three conversion factors build the theoretical yield.

Step 1 · Compute the theoretical yield

Worked example 1 — solution

The balance reads 18.5 g. The first attempt divided by 50.0 g of CaCO₃ — a different substance. The 100% mark is 22.0 g: the most CO₂ this charge can form.

Worked example 1 — solution

Worked example 1 — solution

Worked example 2 — which reactant limits

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

Step 1 · Compute the theoretical yield

Worked example 2 — which reactant limits

Worked example 2 — which reactant limits

Worked example 2 — the percent yield

Step 1 · Compute the theoretical yield

Worked example 2 — the percent yield

Step 1 · Compute the theoretical yield

The dried precipitate weighs 14.2 g. Measured on the balance, not computed.

Worked example 2 — the percent yield

Step 1 · Compute the theoretical yield

Worked example 2 — the percent yield

Step 1 · Compute the theoretical yield

Practice 1 — answer: B

A compared product to starting material: 15.6/40.0 × 100 = 39.0. C left the fraction as a decimal: 15.6/18.8 = 0.830. D skipped the mole ratio in the theoretical yield: 40.0/34.02 × 32.00 = 37.6 g, then 15.6/37.6 × 100 = 41.5.

Practice 1 — answer: B

Worked example 3 — solution

Four conversion factors are needed: the percent yield, then the mass-to-mass chain run from product to reactant.

Theoretical yield required

At 78.0% yield, every 100 g theoretical delivers 78.0 g actual. Two orientations exist. Only one cancels the given g actual:

Worked example 3 — solution

Worked example 3 — solution

Worked example 3 — solution

Practice 2 — answer: A

Practice 2 — answer: A

Practice 2 — answer: A