Hydrogen gas is made by adding zinc to excess dilute hydrochloric acid.
Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)\mathrm{Zn(s) + 2HCl(aq) \rightarrow ZnCl_{2}(aq) + H_{2}(g)}Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)
One mole of any gas occupies 24 dm324\ \mathrm{dm^{3}}24 dm3 at room temperature and pressure.
Relative atomic mass: Zn=65\mathrm{Zn} = 65Zn=65
Calculate the amount, in mol, of hydrogen in 2.4 dm32.4\ \mathrm{dm^{3}}2.4 dm3 of the gas at room temperature and pressure.
Use your answer to (a) and the balanced equation to work out the mass, in g, of zinc required.
A reaction vessel contains 18 cm318\ \mathrm{cm^{3}}18 cm3 of hydrogen and 25 cm325\ \mathrm{cm^{3}}25 cm3 of chlorine.
H2(g)+Cl2(g)→2HCl(g)\mathrm{H_{2}(g) + Cl_{2}(g) \rightarrow 2HCl(g)}H2(g)+Cl2(g)→2HCl(g)
The gases react completely. All gas volumes are measured at the same temperature and pressure.
Calculate the volume of chlorine that reacts with the hydrogen.
Calculate the three values missing from the final-mixture table.
| Gas | Volume after the reaction / cm3\mathrm{cm^{3}}cm3 |
|---|---|
| hydrogen | |
| chlorine | |
| hydrogen chloride |
151 exam-style questions on Edexcel GCSE Chemistry 6.2 Quantitative analysis, covering 6.2.1 Concentration in mol dm⁻³ and titration calculations, 6.2.2 Percentage yield and why actual yield falls short, 6.2.3 Atom economy of a reaction, 6.2.4 Choosing a reaction pathway, and 6.2.5 Molar volume of gases and Avogadro's law. Each one has a worked solution and a mark scheme showing where the marks go.