Difficulty: Intermediate | Prerequisites: Balancing chemical equations, molar mass calculations, understanding of the mole concept.
Stoichiometry is the backbone of quantitative chemistry: it tells you how much of one substance you need (or get) when another reacts. This worksheet focuses on mixed-unit conversions, where the given quantity and the answer are in different units (grams, litres, moles, particles). You should already be comfortable balancing equations and calculating molar masses. If mole-to-mole conversions still feel shaky, revisit those first, because every problem here layers a unit conversion on top of the mole ratio.
Every stoichiometry problem follows the same route: convert what you are given into moles, use the mole ratio from the balanced equation to cross to the other substance, then convert out of moles into whatever unit the question asks for. The only thing that changes from problem to problem is which conversion factor you use on the way in and the way out (molar mass for grams, 22.4 L/mol for litres at STP, Avogadro's number for particles).
Mole (mol)
The SI unit for "amount of substance." One mole contains exactly 6.02 × 10²³ representative particles (atoms, molecules, or formula units). Think of it as the chemist's counting unit, the way a dozen means 12.
Molar mass
The mass of one mole of a substance, expressed in grams per mole (g/mol). You find it by adding up the atomic masses of every atom in the formula. In simple terms, it is the bridge between grams and moles.
Avogadro's number (6.02 × 10²³)
The number of particles in one mole of any substance. It lets you convert between moles and individual atoms or molecules. Think of it as the "conversion rate" from the invisible world of particles to a quantity you can count.
Molar volume at STP (22.4 L/mol)
At standard temperature and pressure (0 °C, 1 atm), one mole of any ideal gas occupies 22.4 litres. This is the bridge between litres of gas and moles, but only at STP.
Stoichiometric ratio (mole ratio)
The ratio of coefficients in a balanced chemical equation. It tells you how many moles of one substance react with or produce moles of another. In simple terms, it is the recipe that says "for every 1 of A, you need 2 of B."
Balanced equation
A chemical equation in which the number of atoms of each element is the same on both sides. Balancing is not optional: the coefficients are the mole ratios you need for every calculation.
Conversion factor
A fraction equal to 1, built from two equivalent quantities (e.g. 1 mol / 22.4 L), used to cancel one unit and introduce another. Every stoichiometry problem is a chain of conversion factors.
Every mixed stoichiometry problem follows the same three-step path:
Convert the given quantity into moles of the given substance. Use molar mass (g → mol), molar volume at STP (L → mol), or Avogadro's number (particles → mol).
Use the mole ratio from the balanced equation to cross from the given substance to the target substance. The coefficients in the balanced equation give you this ratio directly.
Convert moles of the target substance into the requested unit. Again, pick the right bridge: molar mass (mol → g), molar volume (mol → L), or Avogadro's number (mol → particles).
The flowchart in the original worksheet maps this out visually. On the left side sits Substance A (given) with branches to its mass, volume, and particle count. On the right sits Substance B (target) with the same branches. Moles sit in the centre, connected by the coefficient ratio. Every path runs through the centre.
Problem: CO₂ exhaled by a space shuttle crew is removed by reacting with LiOH, producing Li₂CO₃ and H₂O. If a crew member exhales 450 L of CO₂ in a day, what mass of LiOH is needed?
Step 1 – Balance the equation: CO₂ + 2LiOH → Li₂CO₃ + H₂O
Step 2 – Identify the conversion: From CO₂ (litres) to LiOH (grams).
Step 3 – Build the conversion chain:
Convert 450 L CO₂ → moles CO₂: 450 L ÷ 22.4 L/mol = 20.09 mol CO₂
Mole ratio: 1 mol CO₂ requires 2 mol LiOH, so 20.09 × 2 = 40.18 mol LiOH
Convert moles LiOH → grams: molar mass of LiOH = 7 + 16 + 1 = 24 g/mol, so 40.18 × 24 = 964.3 g LiOH
Answer: 964.3 g of LiOH.
Problem: 2NO₂ → N₂ + 2O₂. The engine releases 20.0 L of NO₂ in an hour. How many moles of N₂ come out?
20.0 L NO₂ ÷ 22.4 L/mol = 0.893 mol NO₂
Mole ratio: 2 mol NO₂ produces 1 mol N₂, so 0.893 ÷ 2 = 0.446 mol N₂
Answer: approximately 0.45 mol N₂.
Problem: CH₄ + 2O₂ → CO₂ + 2H₂O. A furnace burns 1.8 g of methane per hour. How many molecules of CO₂ are released?
Molar mass of CH₄ = 12 + 4(1) = 16 g/mol
1.8 g ÷ 16 g/mol = 0.1125 mol CH₄
Mole ratio: 1 mol CH₄ produces 1 mol CO₂, so 0.1125 mol CO₂
0.1125 × 6.02 × 10²³ = 6.77 × 10²² molecules CO₂
Answer: approximately 6.77 × 10²² molecules of CO₂.
Problem: Magnesium reacts with water to produce magnesium hydroxide and hydrogen gas. What mass of water is needed to react with 8.1 g of magnesium?
Balanced equation: Mg + 2H₂O → Mg(OH)₂ + H₂
Molar mass of Mg = 24.3 g/mol
8.1 g ÷ 24.3 g/mol = 0.333 mol Mg
Mole ratio: 1 mol Mg requires 2 mol H₂O, so 0.333 × 2 = 0.667 mol H₂O
Molar mass of H₂O = 18 g/mol, so 0.667 × 18 = 12.0 g H₂O
Answer: 12.0 g of water.
Problem: Methanol (CH₃OH) is synthesised from CO and H₂. If you have 73.6 L of CO, what mass of methanol can you produce?
Balanced equation: CO + 2H₂ → CH₃OH
73.6 L CO ÷ 22.4 L/mol = 3.286 mol CO
Mole ratio: 1 mol CO produces 1 mol CH₃OH, so 3.286 mol CH₃OH
Molar mass of CH₃OH = 12 + 3(1) + 16 + 1 = 32 g/mol, so 3.286 × 32 = 105.1 g
Answer: 105.1 g of methanol.
Problem: Calcium carbonate is heated to produce quicklime (CaO) and CO₂. What mass of quicklime comes from 4,500 g of CaCO₃?
Balanced equation: CaCO₃ → CaO + CO₂
Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100 g/mol
4,500 g ÷ 100 g/mol = 45 mol CaCO₃
Mole ratio: 1:1, so 45 mol CaO
Molar mass of CaO = 40 + 16 = 56 g/mol, so 45 × 56 = 2,520 g
Answer: 2,520 g of quicklime.
Problem: N₂ + 3H₂ → 2NH₃. How many moles of nitrogen are needed to fully react with 25.0 L of hydrogen gas?
25.0 L H₂ ÷ 22.4 L/mol = 1.116 mol H₂
Mole ratio: 3 mol H₂ requires 1 mol N₂, so 1.116 ÷ 3 = 0.372 mol N₂
Answer: approximately 0.37 mol N₂.
Problem: You eat 10 g of glucose (C₆H₁₂O₆). How many litres of CO₂ will you exhale from its combustion in your cells?
Balanced equation: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
Molar mass of C₆H₁₂O₆ = 6(12) + 12(1) + 6(16) = 180 g/mol
10 g ÷ 180 g/mol = 0.0556 mol glucose
Mole ratio: 1 mol glucose produces 6 mol CO₂, so 0.0556 × 6 = 0.333 mol CO₂
0.333 × 22.4 L/mol = 7.47 L CO₂
Answer: approximately 7.5 L of CO₂ at STP.
\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\text{moles} = \frac{\text{volume (L)}}{22.4 \text{ L/mol}} \quad (\text{at STP only})\text{moles} = \frac{\text{number of particles}}{6.02 \times 10^{23}}\text{moles of B} = \text{moles of A} \times \frac{\text{coefficient of B}}{\text{coefficient of A}}These four relationships are all you need. Every mixed stoichiometry problem uses some combination of them, chained together. The mole ratio always sits in the middle of the chain.
Stoichiometry is how engineers size chemical reactors, how pharmacists calculate drug dosages, and how environmental scientists estimate pollutant output from combustion. Problem 1 (catalytic converter) and Problem 7 (cellular respiration of glucose) are direct, everyday examples: your car's exhaust system and your own metabolism both depend on the same mole-ratio logic covered here.
Students often use 22.4 L/mol for liquids or solids. It applies only to gases at STP. If the substance is not a gas, or conditions are not at STP, this conversion factor does not work.
Students frequently forget to balance the equation first and then use incorrect mole ratios. The coefficients are the ratios. An unbalanced equation gives the wrong answer every time.
A common error is dividing by the mole ratio when you should multiply (or vice versa). Always check: are you going from the substance with the larger coefficient to the smaller, or the other way round? Set it up so the units cancel properly.
Some students confuse molar mass with molecular mass or atomic mass number. Molar mass has units of g/mol and is numerically equal to the molecular mass in amu, but the unit matters for dimensional analysis.
⚠️ Mixed-unit stoichiometry problems (given grams, asked for litres, etc.) are among the most commonly tested question types in introductory chemistry exams. Expect at least one on every test.
⚠️ Professors frequently test whether you can write and balance the equation yourself, not just use a given one. Problems 3 through 7 on this worksheet require you to do exactly that.
⚠️ Dimensional analysis (showing that units cancel correctly across the chain) is often required for full marks, even if your numerical answer is correct.
⚠️ The 22.4 L/mol figure is a favourite source of trick questions. If the problem states conditions other than STP, you need the ideal gas law (PV = nRT) instead.
True or false: 22.4 L/mol can be used to convert litres of liquid water to moles. Answer: False. 22.4 L/mol applies only to gases at STP.
Fill in the blank: The coefficients in a balanced equation give you the ______ ratio between substances. Answer: mole
True or false: You must always convert to moles before you can cross from one substance to another using the balanced equation. Answer: True.
Fill in the blank: The molar mass of CO₂ is ______ g/mol. Answer: 44 (12 + 2 × 16)
True or false: If a reaction has a 1:1 mole ratio, the mass of reactant consumed always equals the mass of product formed. Answer: False. A 1:1 mole ratio means equal moles, not equal masses. The masses differ unless both substances happen to have the same molar mass.
Q: If 5.0 g of NaOH reacts with excess HCl (NaOH + HCl → NaCl + H₂O), how many grams of NaCl are produced?
A: Molar mass of NaOH = 40 g/mol. 5.0 ÷ 40 = 0.125 mol NaOH. The mole ratio is 1:1, so 0.125 mol NaCl. Molar mass of NaCl = 58.5 g/mol. 0.125 × 58.5 = 7.3 g NaCl.
Q: How many litres of O₂ at STP are needed to completely combust 3.0 mol of propane (C₃H₈ + 5O₂ → 3CO₂ + 4H₂O)?
A: 3.0 mol C₃H₈ × 5 mol O₂ / 1 mol C₃H₈ = 15 mol O₂. 15 × 22.4 = 336 L O₂.
Q: In the reaction 2Al + 3Cl₂ → 2AlCl₃, how many molecules of Cl₂ are needed to react with 5.4 g of aluminium?
A: Molar mass of Al = 27 g/mol. 5.4 ÷ 27 = 0.2 mol Al. Mole ratio: 2 mol Al requires 3 mol Cl₂, so 0.2 × (3/2) = 0.3 mol Cl₂. 0.3 × 6.02 × 10²³ = 1.81 × 10²³ molecules.
Q: A student is given 44.8 L of N₂ at STP. Using the Haber process (N₂ + 3H₂ → 2NH₃), what mass of ammonia could be produced?
A: 44.8 ÷ 22.4 = 2.0 mol N₂. Mole ratio: 1 mol N₂ produces 2 mol NH₃, so 4.0 mol NH₃. Molar mass of NH₃ = 17 g/mol. 4.0 × 17 = 68 g NH₃.
This material connects directly to limiting reagent problems, which add one more step: when you have amounts of two reactants, you run this same conversion for each and determine which runs out first. It also leads into gas law stoichiometry (using PV = nRT instead of 22.4 L/mol when conditions are not at STP) and solution stoichiometry (using molarity × volume instead of molar mass to get moles). If you are comfortable with the flowchart method here, those extensions are a straightforward bolt-on.
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