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Chemical formulae, equations and calculations

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Question 1

Nitric acid can be neutralised by adding sodium hydroxide solution. The equation for the reaction is:

HNO3+NaOH→NaNO3+H2O \text{HNO}_3 + \text{NaOH} \rightarrow \text{NaNO}_3 + \text{H}_2\text{O} HNO3​+NaOH→NaNO3​+H2​O

A solution of nitric acid has a concentration of 0.240 mol/dm30.240\text{ mol/dm}^30.240 mol/dm3.

a.

Calculate the amount, in moles, of HNO3\text{HNO}_3HNO3​ in 75.0 cm375.0\text{ cm}^375.0 cm3 of the nitric acid solution.

[2]
b.

Calculate the volume of 0.150 mol/dm30.150\text{ mol/dm}^30.150 mol/dm3 sodium hydroxide solution needed to exactly neutralise the nitric acid. Give the unit.

[3]
c.

In another neutralisation reaction, a student uses 30.0 cm330.0\text{ cm}^330.0 cm3 of 0.800 mol/dm30.800\text{ mol/dm}^30.800 mol/dm3 aqueous sodium hydroxide solution.

Calculate the mass of sodium hydroxide contained in this solution. [Relative atomic masses, ArA_rAr​: H=1\text{H} = 1H=1, O=16\text{O} = 16O=16, Na=23\text{Na} = 23Na=23]

[3]

Chemical formulae, equations and calculations Questions

  1. IGCSE
  2. /Chemistry
  3. /Chemical formulae, equations and calculations