Reacting masses and volumes (of solutions and gases)

AS Level Chemistry · Atoms, molecules and stoichiometry. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

A balanced equation gives the ratio in which substances react, counted in moles. Every calculation follows the same path: change what you are given into moles, use the ratio in the equation, then change the moles of the new substance into the mass, gas volume or concentration you want.

If amounts of two reactants are given, one may run out first. This is the limiting reagent, and it decides how much product forms. The other reactant is in excess.

The percentage yield compares what you actually collect with the maximum the equation allows. Give answers to the number of significant figures in the data.

Rules to remember

Common mistake

Students use the reactant in excess to work out the product, or forget to change cm3 to dm3. Find the limiting reagent first, and divide cm3 by 1000.

Worked example

3.27 g of zinc (Ar = 65.4) is added to 50.0 cm3 of 0.500 mol dm−3 copper(II) sulfate. Zn + CuSO4 → ZnSO4 + Cu. What mass of copper (Ar = 63.5) forms?

  1. Moles of Zn = 3.27 ÷ 65.4 = 0.0500 mol.
  2. Moles of CuSO4 = 0.500 × (50.0 ÷ 1000) = 0.0250 mol.
  3. The ratio is 1 : 1, so CuSO4 is limiting and 0.0250 mol of Cu forms.
  4. Mass of Cu = 0.0250 × 63.5 = 1.59 g.

Answer: 1.59 g of copper (zinc is in excess).

Practice questions

Try each one, then open the answer.

1. 20 cm3 of ethane is exploded with 100 cm3 of oxygen (an excess), and the mixture is cooled to room temperature. What is the total volume of gas remaining?

  1. A
    40 cm3
  2. B
    70 cm3
  3. C
    100 cm3
  4. D
    130 cm3
Show answer

Answer: B. C2H6 + 3½O2 → 2CO2 + 3H2O. 20 cm3 of ethane uses 70 cm3 of O2 and forms 40 cm3 of CO2; the water condenses. Gas left = 30 cm3 O2 + 40 cm3 CO2 = 70 cm3.

2. 10.0 g of calcium carbonate (Mr = 100.1) is heated strongly and 4.48 g of calcium oxide (Mr = 56.1) is collected. What is the percentage yield?

  1. A
    44.8%
  2. B
    79.9%
  3. C
    56.0%
  4. D
    89.6%
Show answer

Answer: B. Moles CaCO3 = 10.0/100.1 = 0.0999; theoretical CaO = 0.0999 × 56.1 = 5.60 g. Yield = 4.48/5.60 × 100 = 79.9%. Dividing 4.48 by the 10.0 g of starting material gives the wrong 44.8%.

3. What mass of sodium hydroxide is needed to make 250 cm3 of a 0.200 mol dm−3 solution?

  1. A
    8.00 g
  2. B
    0.200 g
  3. C
    10.0 g
  4. D
    2.00 g
Show answer

Answer: D. n = 0.200 × 250/1000 = 0.0500 mol; Mr of NaOH = 23.0 + 16.0 + 1.0 = 40.0; mass = 0.0500 × 40.0 = 2.00 g. 8.00 g would make 1 dm3.

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