5PR - Step-by-step solutions to SRP Questions (Molar Calculations)

A set of chemical calculations involving the mole and chemical equations are given on 12 Feb to 5PR during Structured Revision Programme.

Please study the steps taken to solve each of the 4 problems from the section on Calculations From Equations (p4 to 6)

Q1(a) No. of mole of iron used = mass of iron used / atomic mass of iron = 28/56 = 0.5 mol

From the balanced equation, 1 mole of iron produces 1 mole of iron(II) chloride.

Hence, 0.5 mol of iron used produces 0.5 mol of iron(II) chloride.

To find mass of iron(II) chloride formed,

Molar mass of iron (II) chloride = 56 + (35.5 x 2) = 127g. (this means mass of 1 mole of iron(II) chloride is 127g)

Mass of iron(II) chloride found in 0.5 mol = 0.5 x 127g = 63.5g

Q1(b) Use Avogadro's Law (1 mole of any gas occupies a volume of 24 dm3 at room temperature and pressure (r.t.p.)) to solve this problem.

From the equation, 1 mole of iron produces 1 mole of hydrogen gas.

Therefore, 0.5 mol of iron (from 1(a)) produces 0.5 mol of hydrogen at r.t.p.

Volume of hydrogen produced = 0.5 x 24 = 12 dm3.


Q2. Step 1: No. of moles of CuCO3 heated = 74.4 / (molar mass of CuCO3)
= 74.4 / (64 + 12 + 48)
= 0.6 mol

Step 2: State that 1 mole of CuCO3 decomposes on heating to produce 1 mole of CuO.
Hence, 0.6 mol of CuO is produced from heating 0.6 mol of CuCO3 (from step 1).

Step 3: Therefore, mass of CuO (in 0.6 mol) = 0.6 x (molar mass of CuO)
= 0.6 x (64 + 16)
= 48 g


Q3. Same problem solving method as Q1b.

No. of moles of HCl reacted (14.8 g) = 14.8 / (1 + 35.5) = 0.406 mol.

From the equation, 2 moles of HCl reacts to produce 1 mole of sulphur dioxide (SO2) gas.

Hence, if 0.406 mol of HCl has reacted, 0.406 / 2 = 0.203 mol of SO2 gas is produced.

Using Avogadro's Law, volume of SO2 produced at r.t.p. = 0.203 x 24 dm3

= 4.87 dm3 (3 s.f.)


Q4. Same problem solving method as Q2.

Work out number of mole of copper in the 10.8g used (0.169 mol)

Find number of mole of aluminium produced from the chemical equation (0.113 mol)

Work out mass of aluminium solid in 0.113 mol (3.04 g) [Atomic mass of Al is 27].


Key point - You must convert all mass (in g or kg) and volume (in cm3 or dm3) into the equivalent values in mole (mol) to use the chemical equation to compare and work out respective amounts.