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K324 / 2027
Chemical Calculations

Lesson 4 of 6

Use the equation ratio

Convert to moles before comparing reactants.

O-Level 6092 (2026) / SEC G3 K324 (2027)

  • I can interpret equations and states.Syllabus

    Syllabus K324, 4.1(d). Interpret equations and states

  • I can calculate reacting masses and gas volumes.Syllabus

    Syllabus K324, 4.2(f). Calculate reacting masses and gas volumes

Use the equation ratio

Convert to moles before comparing reactants.

Make a guess. You are not marked.

3.60 g of magnesium burns completely in oxygen to make magnesium oxide. How does the mass of magnesium oxide compare with 3.60 g?

  1. It is less than 3.60 g, as some burns away
  2. It is exactly 3.60 g, as mass is conserved
  3. It is more than 3.60 g, as oxygen is added
Show the answer

It is more than 3.60 g, as oxygen is added

The oxide holds all the magnesium plus the oxygen it took from the air, so the product is heavier. The worked example below finds 6.00 g.

A reliable calculation path
  1. Write and balance the reaction

    The coefficients give a mole ratio, not a mass ratio.

  2. Convert the known quantity to moles

    Mass: n = m/M. Gas at room temperature and pressure: n = V/24 with V in dm3.

  3. Apply the mole ratio

    For aA -> bB, n(B) = n(A) x b/a.

  4. Convert to the requested quantity

    Use m = nM or V = 24n at room temperature and pressure.

Worked exampleMass to gas volume

Excess hydrochloric acid reacts with 2.40 g Mg. Calculate hydrogen volume at room temperature and pressure; Mg = 24.

Try it first. Then show the worked stepsHide the worked steps
  1. Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g).
  2. n(Mg) = 2.40/24 = 0.100 mol; the Mg:H2 ratio is 1:1.
  3. V(H2) = 0.100 x 24 = 2.40 dm3.
Answer

2.40 dm3, or 2400 cm3. The 24 dm3 mol-1 value applies at room temperature and pressure.

Try it step by step

Work it out in steps: Moles of the solid, then Moles of gas from the ratio, then Volume of gas.

  1. Problem 1. Excess dilute sulfuric acid reacts with 1.30 g of zinc. Zn(s) + H2SO4(aq) -> ZnSO4(aq) + H2(g). Find the volume of hydrogen at room temperature and pressure, in dm3 to 3 significant figures. Use Zn = 65.
  2. Problem 2. Excess hydrochloric acid reacts with 5.00 g of calcium carbonate. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g). Find the volume of carbon dioxide at room temperature and pressure, in dm3 to 3 significant figures. Use M(CaCO3) = 100 g mol-1.
  3. Problem 3. Excess hydrochloric acid reacts with 0.540 g of aluminium. 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g). Find the volume of hydrogen at room temperature and pressure. Give each answer to 3 significant figures. Use Al = 27.
Show the answer

Problem 1.

Moles of the solid: n(Zn) = 1.30 / 65 = 0.0200 mol.

Moles of gas from the ratio: The equation has 1 Zn to 1 H2, so n(H2) = 0.0200 mol.

Volume of gas: V = n x 24 = 0.0200 x 24 = 0.480 dm3 at room temperature and pressure.

Problem 2.

Moles of the solid: n(CaCO3) = 5.00 / 100 = 0.0500 mol.

Moles of gas from the ratio: The equation has 1 CaCO3 to 1 CO2, so n(CO2) = 0.0500 mol. The 2 in front of HCl does not matter here, because the acid is in excess.

Volume of gas: V = 0.0500 x 24 = 1.20 dm3.

Problem 3.

Moles of the solid: n(Al) = 0.540 / 27 = 0.0200 mol.

Moles of gas from the ratio: The equation has 2 Al to 3 H2, so n(H2) = 0.0200 x 3/2 = 0.0300 mol.

Volume of gas: V = 0.0300 x 24 = 0.720 dm3.

Worked exampleMass to mass through the mole ratio

What mass of MgO forms when 3.60 g Mg burns completely in excess oxygen? Use Mg = 24 and O = 16.

Try it first. Then show the worked stepsHide the worked steps
  1. 2Mg(s) + O2(g) -> 2MgO(s). The Mg:MgO ratio is 2:2, or 1:1.
  2. n(Mg) = 3.60/24 = 0.150 mol, so n(MgO) = 0.150 mol.
  3. M(MgO) = 24 + 16 = 40 g mol-1; mass = 0.150 x 40 = 6.00 g.
Answer

6.00 g MgO. The extra 2.40 g is oxygen taken from the air; the 1:1 mole ratio does not mean equal masses.

Worked exampleIdentify the limiting reactant

Mix 0.080 mol Mg with 0.100 mol HCl. How much H2 forms?

Try it first. Then show the worked stepsHide the worked steps
  1. The equation needs two moles of HCl for each mole of Mg.
  2. 0.080 mol Mg would require 0.160 mol HCl; only 0.100 mol is available. HCl is limiting.
  3. n(H2) = 0.100/2 = 0.050 mol. Mg used = 0.050 mol; Mg remaining = 0.030 mol.
Answer

0.050 mol hydrogen, equivalent to 1.20 dm3 at room temperature and pressure.

Check your understanding

Can 0.10 mol of Mg react completely with 0.10 mol of HCl?

  1. No. Mg + 2HCl needs 0.20 mol HCl, so HCl runs out and 0.050 mol Mg is left.
  2. No. HCl is in excess, because one mole of acid gives two moles of H+.
  3. Yes. The amounts are equal, so each reactant is used up at the same time.
Show the answer

No. Mg + 2HCl needs 0.20 mol HCl, so HCl runs out and 0.050 mol Mg is left.

The equation needs 2 mol of HCl for each mol of Mg. 0.10 mol HCl reacts with 0.050 mol Mg, so HCl is limiting and 0.050 mol Mg remains.