How to Use This Worksheet
Work in order. Section A checks the ideas: where mole ratios come from, the key definitions, and the basic conversions. Section B builds the skill: mass to mass calculations, limiting reactants, percent yield and solution stoichiometry. Section C combines everything in exam style, including testing claims that sound plausible but fail.
Write the three steps of every calculation out in full: convert to moles, apply the mole ratio from the balanced equation, then convert to what is asked. Carry unrounded values through and round only the final answer, to three significant figures. Molar masses you need: H₂O 18.02, NH₃ 17.03, CO₂ 44.01, NaCl 58.44, CaCO₃ 100.09, KClO₃ 122.55, MgO 40.31, ZnCl₂ 136.28 g/mol; atomic masses: H 1.008, C 12.01, O 16.00, Mg 24.31, Zn 65.38 g/mol.
Two habits separate correct answers from near misses in this topic. First, never compare reactants by mass or by moles alone when deciding which one limits: test how much product each could make. Second, treat a percent yield above 100% as an alarm bell, never as a success. Finally, whenever both products’ masses are known, check them against the reactant masses consumed: conservation of mass catches arithmetic slips.
Do not look at the answer key until you have attempted every question.
Section A: Recall and Concepts
A1.
(a) State where the mole ratio between two substances in a reaction comes from.
(b) For 2H₂ + O₂ → 2H₂O, state how many moles of H₂O form from 3.0 mol of O₂, and how many moles of O₂ are needed for 5.0 mol of H₂.
(c) For N₂ + 3H₂ → 2NH₃, calculate the moles of NH₃ formed from 9.0 mol of N₂ with H₂ in excess.
(d) Explain why the subscripts in a formula cannot be used as a mole ratio.
A2. Define each term in one sentence.
(a) Mole ratio.
(b) Limiting reactant.
(c) Excess reactant.
(d) Theoretical yield.
(e) Actual yield.
(f) Percent yield, including the formula.
A3. Potassium chlorate decomposes: 2KClO₃ → 2KCl + 3O₂.
(a) How many moles of O₂ form from 5.0 mol of KClO₃?
(b) How many moles of KClO₃ are needed to produce 6.0 mol of O₂?
(c) How many moles of KCl form alongside the O₂ in part (a)?
A4.
(a) Calculate the moles in 11.0 g of CO₂.
(b) Calculate the mass of 0.750 mol of NaCl.
(c) Calculate the moles in 90.0 g of H₂O.
(d) State the two conversion factors that connect mass and moles, and what each one does.
A5. Magnesium burns: 2Mg + O₂ → 2MgO.
(a) A mixture holds 4.0 mol of Mg and 4.0 mol of O₂. Which reactant is limiting, and how much O₂ is left over?
(b) A second mixture holds 4.0 mol of Mg and 1.0 mol of O₂. Which reactant is limiting now, how much MgO forms, and how much Mg is left over?
(c) Explain why “the reactant present in the smaller amount” is not a safe way to choose the limiting reactant.
Section B: Core Practice
B1. Methane burns completely: CH₄ + 2O₂ → CO₂ + 2H₂O. A sample of 10.0 g of CH₄ is burned in excess oxygen.
(a) Calculate the moles of CH₄ burned.
(b) Calculate the mass of CO₂ produced.
(c) Calculate the mass of O₂ consumed.
B2. A ribbon of 8.00 g of Mg is burned in 6.00 g of O₂: 2Mg + O₂ → 2MgO.
(a) Calculate the moles of each reactant present.
(b) By calculating the MgO each reactant could produce, identify the limiting reactant.
(c) Calculate the theoretical yield of MgO.
(d) Calculate the mass of the excess reactant left over.
B3.
(a) A reaction has a theoretical yield of 25.5 g and an actual yield of 20.4 g. Calculate the percent yield.
(b) A reaction has a percent yield of 85.0% and a theoretical yield of 40.0 g. Calculate the actual yield.
(c) A student reports a percent yield of 105%. Give two distinct reasons why the isolated product cannot be pure and dry.
(d) In part (a), how much more product would need to be recovered to bring the yield up to 90.0%?
B4. Hydrochloric acid is neutralised by sodium hydroxide: HCl + NaOH → NaCl + H₂O.
(a) Calculate the moles of HCl in 25.0 mL of 0.200 M HCl.
(b) Calculate the volume of 0.150 M NaOH needed to neutralise exactly the acid in part (a).
(c) Sulfuric acid reacts differently: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the moles of NaOH needed to neutralise 20.0 mL of 0.500 M H₂SO₄.
(d) Calculate the volume of 0.400 M NaOH that supplies the moles in part (c).
(e) Calculate the mass of NaCl formed in the neutralisation in parts (a) and (b).
B5. Ammonia is made by N₂ + 3H₂ → 2NH₃. A reactor is charged with 14.0 g of N₂ and 4.00 g of H₂.
(a) Calculate the moles of each reactant.
(b) Identify the limiting reactant, showing the test you used.
(c) Calculate the theoretical yield of NH₃.
(d) The reactor actually produces 12.8 g of NH₃. Calculate the percent yield.
(e) Calculate the mass of H₂ left over in the reactor when the limiting reactant is exhausted.
B6. Zinc reacts with excess hydrochloric acid: Zn + 2HCl → ZnCl₂ + H₂. A 13.1 g piece of zinc is used.
(a) Calculate the moles of zinc.
(b) Calculate the moles of HCl consumed.
(c) Calculate the mass of H₂ produced.
(d) Calculate the mass of ZnCl₂ produced, and check your answers using conservation of mass. (Molar mass of HCl is 36.46 g/mol.)
Section C: Exam Style
C1. A student adds 20.0 g of CaCO₃ to 100 mL of 3.00 M HCl: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂.
(a) Calculate the moles of each reactant present.
(b) Identify the limiting reactant, showing your working.
(c) Calculate the theoretical yield of CO₂.
(d) The student collects 5.94 g of CO₂. Calculate the percent yield.
(e) Calculate the mass of CaCO₃ left unreacted.
C2. Propane burns: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. A burner is supplied with 11.0 g of C₃H₈ and 30.0 g of O₂.
(a) Calculate the moles of each reactant.
(b) Identify the limiting reactant, showing your working.
(c) Calculate the masses of CO₂ and of H₂O produced.
(d) Calculate the mass of the excess reactant left over.
C3. Three students make claims about the reaction 2Mg + O₂ → 2MgO. Evaluate each claim.
Claim 1: “With 8.00 g of Mg and 6.00 g of O₂, the oxygen must be the limiting reactant, because it has the smaller mass.”
Claim 2: “My percent yield was 110%, so my reaction worked better than the theory predicted.”
Claim 3: “If I mix exactly 2.0 mol of Mg with 2.0 mol of O₂, neither reactant is in excess, because the amounts are equal.”
(a) Evaluate Claim 1 by carrying out the limiting reactant test.
(b) Evaluate Claim 2, explaining what a yield above 100% really indicates.
(c) Evaluate Claim 3 by working out what actually happens to each reactant.
Show the answer key and full working (only after you have attempted every question)
A1. (a) From the coefficients of the balanced equation: the ratio of the coefficients of the two substances. (b) 6.0 mol of H₂O (3.0 × 2/1), and 2.5 mol of O₂ (5.0 × 1/2). (c) 9.0 × 2/1 = 18.0 mol of NH₃. (d) Subscripts state the fixed composition of one formula unit, for example the two H atoms inside each H₂O molecule. They describe what a substance is, not how much of it reacts; only the coefficients in front of the formulas give reacting amounts.
A2. (a) The ratio between the amounts in moles of any two substances in a reaction, taken from the coefficients of the balanced equation. (b) The reactant that is completely consumed first, stopping the reaction and setting the maximum amount of product. (c) A reactant supplied in more than the amount the limiting reactant can consume, so some remains when the reaction stops. (d) The maximum amount of product predicted by the stoichiometric calculation from the limiting reactant, assuming perfect conversion and no losses. (e) The amount of product actually obtained and isolated in a real experiment. (f) Percent yield = (actual yield ÷ theoretical yield) × 100.
A3. (a) 5.0 × 3/2 = 7.5 mol of O₂. (b) 6.0 × 2/3 = 4.0 mol of KClO₃. (c) The KClO₃ to KCl ratio is 2 : 2, so 5.0 mol of KCl forms.
A4. (a) 11.0 ÷ 44.01 = 0.250 mol. (b) 0.750 × 58.44 = 43.8 g. (c) 90.0 ÷ 18.02 = 4.99 mol, essentially 5.00 mol. (d) Dividing a mass by the molar mass converts grams into moles; multiplying moles by the molar mass converts back into grams. The molar mass is the bridge in both directions.
A5. (a) Magnesium. It needs 4.0 ÷ 2 = 2.0 mol of O₂, so 4.0 − 2.0 = 2.0 mol of O₂ is left over. (b) Oxygen. It consumes 2 × 1.0 = 2.0 mol of Mg and makes 2.0 mol of MgO, leaving 4.0 − 2.0 = 2.0 mol of Mg. (c) Because the equation may demand the reactants in an unequal ratio. Safety comes only from comparing the amounts through the coefficients, or equivalently from testing how much product each reactant could make.
B1. (a) 10.0 ÷ 16.04 = 0.623 mol of CH₄ (molar mass of CH₄ = 12.01 + 4 × 1.008 = 16.04 g/mol). (b) The ratio is 1 : 1, so 0.623 mol of CO₂, with mass 0.623 × 44.01 = 27.4 g. (c) O₂ consumed = 2 × 0.623 = 1.25 mol, with mass 1.25 × 32.00 = 39.9 g.
B2. (a) Mg: 8.00 ÷ 24.31 = 0.329 mol. O₂: 6.00 ÷ 32.00 = 0.188 mol. (b) From the Mg: 0.329 mol of MgO. From the O₂: 0.188 × 2 = 0.375 mol of MgO. Magnesium produces less, so magnesium is limiting, even though the O₂ had the smaller mass and fewer moles. (c) 0.329 × 40.31 = 13.3 g of MgO. (d) O₂ consumed = 0.329 ÷ 2 = 0.165 mol, so 0.188 − 0.165 = 0.0230 mol remains, with mass 0.0230 × 32.00 = 0.735 g.
B3. (a) 20.4 ÷ 25.5 × 100 = 80.0%. (b) 0.850 × 40.0 = 34.0 g. (c) Any two of: the product is still wet with solvent or water, adding mass that is not product; the product is contaminated with unreacted reactant or a side product; the solid was not fully dried before weighing. A weighing or recording error is also possible. A genuine yield of pure product cannot exceed the theoretical maximum. (d) A 90.0% yield needs 0.900 × 25.5 = 22.95 g, so 22.95 − 20.4 = 2.55 g more product.
B4. (a) 0.0250 L × 0.200 = 0.00500 mol of HCl. (b) The ratio is 1 : 1, so 0.00500 mol of NaOH is needed: volume = 0.00500 ÷ 0.150 = 0.0333 L = 33.3 mL. (c) Moles of H₂SO₄ = 0.0200 × 0.500 = 0.0100 mol. The ratio is 1 : 2, so NaOH needed = 2 × 0.0100 = 0.0200 mol. (d) Volume = 0.0200 ÷ 0.400 = 0.0500 L = 50.0 mL. (e) NaCl formed = 0.00500 mol (the 1 : 1 ratio again), with mass 0.00500 × 58.44 = 0.292 g.
B5. (a) N₂: 14.0 ÷ 28.02 = 0.500 mol. H₂: 4.00 ÷ 2.016 = 1.98 mol. (b) Product from the N₂: 2 × 0.500 = 1.00 mol of NH₃. Product from the H₂: 1.98 × 2/3 = 1.32 mol of NH₃. Nitrogen makes less, so N₂ is limiting. (Equivalently, 0.500 mol of N₂ needs 1.50 mol of H₂, and 1.98 mol is available.) (c) 1.00 mol × 17.03 = 17.0 g of NH₃. (d) 12.8 ÷ 17.0 × 100 = 75.2%. (e) H₂ consumed = 1.50 mol, leaving 1.98 − 1.50 = 0.485 mol, with mass 0.485 × 2.016 = 0.978 g.
B6. (a) 13.1 ÷ 65.38 = 0.200 mol of Zn. (b) The ratio is 1 : 2, so HCl consumed = 2 × 0.200 = 0.401 mol. (c) H₂ formed = 0.200 mol, with mass 0.200 × 2.016 = 0.404 g. (d) ZnCl₂ formed = 0.200 mol, with mass 0.200 × 136.28 = 27.3 g. Check: reactants used = 13.1 g of Zn + 0.401 × 36.46 = 14.6 g of HCl, totalling 27.7 g. Products = 27.3 + 0.404 = 27.7 g. The totals agree, as conservation of mass requires.
C1. (a) CaCO₃: 20.0 ÷ 100.09 = 0.200 mol. HCl: 0.100 L × 3.00 = 0.300 mol. (b) The 0.200 mol of CaCO₃ would need 2 × 0.200 = 0.400 mol of HCl, but only 0.300 mol is present, so HCl is limiting. (Equally, product from HCl is 0.150 mol of CO₂ against 0.200 mol from the carbonate.) (c) CO₂ = 0.300 ÷ 2 = 0.150 mol, with mass 0.150 × 44.01 = 6.60 g. (d) 5.94 ÷ 6.60 × 100 = 90.0%. (e) CaCO₃ consumed = 0.150 mol, so 0.200 − 0.150 = 0.0498 mol remains, with mass 0.0498 × 100.09 = 4.99 g.
C2. (a) C₃H₈: 11.0 ÷ 44.09 = 0.249 mol (molar mass 3 × 12.01 + 8 × 1.008 = 44.09 g/mol). O₂: 30.0 ÷ 32.00 = 0.938 mol. (b) Burning all the propane needs 5 × 0.249 = 1.25 mol of O₂, but only 0.938 mol is present, so O₂ is limiting. (Equally, the O₂ can burn 0.938 ÷ 5 = 0.188 mol of propane.) (c) CO₂ = 3 × 0.188 = 0.563 mol, with mass 0.563 × 44.01 = 24.8 g. H₂O = 4 × 0.188 = 0.750 mol, with mass 0.750 × 18.02 = 13.5 g. (d) Propane consumed = 0.188 mol, leaving 0.249 − 0.188 = 0.0619 mol, with mass 0.0619 × 44.09 = 2.73 g.
C3. (a) Claim 1 is wrong. Moles present: Mg = 8.00 ÷ 24.31 = 0.329 mol; O₂ = 6.00 ÷ 32.00 = 0.188 mol. MgO possible from the Mg is 0.329 mol; from the O₂ it is 0.188 × 2 = 0.375 mol. Magnesium makes less product, so magnesium is limiting despite its larger mass. Mass alone, without moles and coefficients, cannot identify the limiting reactant. (b) Claim 2 is wrong. The theoretical yield is the maximum possible mass of pure product, so 110% cannot be achieved by any improvement in technique. It means the weighed material was not pure dry product, for example still wet or contaminated, or that the theoretical yield was miscalculated, perhaps from the excess reactant. (c) Claim 3 is wrong. The equation consumes Mg and O₂ in a 2 : 1 mole ratio, not 1 : 1. The 2.0 mol of Mg consumes only 1.0 mol of O₂ and makes 2.0 mol of MgO; 1.0 mol of O₂ is left over, so oxygen was in excess even though the starting moles were equal.