A balanced equation gives a ratio of moles, not of grams. Change what you know into moles, use the ratio, change back. When one reactant runs out first it decides how much product you get, and real reactions rarely give all of it. Extended content.
Use three steps: (1) change the mass (or gas volume) you are given into moles; (2) use the numbers in the balanced equation to find the moles of the substance you want; (3) change those moles back into a mass or volume. The limiting reactant is the one that is completely used up; it fixes the amount of product. Percentage yield = actual mass ÷ expected mass × 100. Percentage purity = mass of pure substance ÷ mass of impure sample × 100.
Fig. 1: the three-step route from mass of A to mass of B
Section 01
Three steps through the mole.
Extended content. Always start from a balanced equation. The numbers in front give the ratio of moles, so the masses themselves are never compared directly. Use Ar: C 12, O 16, Mg 24, Ca 40.
Heating limestone: CaCO3 → CaO + CO2. 50 g of CaCO3 is 50 ÷ 100 = 0.5 mol, which gives 0.5 mol of CaO: 0.5 × 56 = 28 g.
The same reaction, other product: 10 g of CaCO3 = 0.1 mol, giving 0.1 mol of CO2 = 0.1 × 44 = 4.4 g (or 0.1 × 24 = 2.4 dm3 at r.t.p.).
Burning magnesium: 2Mg + O2 → 2MgO. 4.8 g of Mg = 4.8 ÷ 24 = 0.2 mol. The ratio Mg : MgO is 2 : 2, so 0.2 mol of MgO = 0.2 × 40 = 8.0 g.
Examiner trap
The 2 in front of 2MgO belongs in the mole ratio, not in the Mr. Use Mr(MgO) = 40, not 80. A student who calculates 0.2 mol × 80 gets 16 g, double the correct 8.0 g.
Section 02
The limiting reactant.
Set the amounts of the two reactants in the simulator and watch which one runs out first. The product stops forming the moment the limiting reactant is used up; the other reactant is left over (in excess).
The reaction: Mg + 2HCl → MgCl2 + H2, with 0.1 mol of Mg and 0.1 mol of HCl.
Check the ratio: 0.1 mol of Mg needs 2 × 0.1 = 0.2 mol of HCl, but only 0.1 mol is there. So HCl is limiting and Mg is in excess.
Product from the limiting reactant: 2 mol HCl gives 1 mol H2, so 0.1 mol HCl gives 0.05 mol H2 = 0.05 × 24 = 1.2 dm3 at r.t.p.
Left over: only 0.05 mol of Mg reacts, so 0.05 mol of Mg remains.
Examiner trap
Equal numbers of moles do not mean nothing is left over. Here there are 0.1 mol of each, but the equation needs twice as much HCl as Mg, so HCl runs out first. Always compare the moles you have with the ratio in the equation.
Stage 1 · Learn
Check what the sim just showed you
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
What mass of calcium oxide forms when 50 g of calcium carbonate is heated? CaCO3 → CaO + CO2 (Mr: CaCO3 100, CaO 56)
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In 2Mg + O2 → 2MgO, how many moles of MgO form from 0.2 mol of Mg?
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The limiting reactant is the reactant that:
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Percentage yield is calculated as:
Section 03
Yield and purity.
Try the yield tab: real reactions lose product on the way, and real starting materials contain impurities.
Percentage yield: 50 g of CaCO3 should give 28 g of CaO. If 21 g is collected, the yield is 21 ÷ 28 × 100 = 75%.
Percentage purity: 5.0 g of impure CaCO3 gives 1.76 g of CO2 = 1.76 ÷ 44 = 0.04 mol. That came from 0.04 mol of CaCO3 = 0.04 × 100 = 4.0 g. Purity = 4.0 ÷ 5.0 × 100 = 80%.
Why yields are below 100%: the reaction may not go to completion, some product is lost when filtering or pouring, and side reactions can make other products.
Stage 2 · Exam
Exam-style questions
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What mass of carbon dioxide is produced when 10 g of calcium carbonate is heated? CaCO3 → CaO + CO2 (Ar: C 12, O 16, Ca 40)
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What mass of magnesium oxide forms when 4.8 g of magnesium burns completely? 2Mg + O2 → 2MgO (Ar: O 16, Mg 24)
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0.1 mol of magnesium is added to 0.1 mol of hydrochloric acid. Mg + 2HCl → MgCl2 + H2. What volume of hydrogen forms at r.t.p.?
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5.0 g of impure calcium carbonate reacts with excess acid to give 1.76 g of carbon dioxide. What is the percentage purity of the calcium carbonate? (Mr: CaCO3 100, CO2 44)
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