Hydrogen reacts with oxygen to form water. The reaction is represented by the equation:
2H2(g)+O2(g)→2H2O(l)2\mathrm{H_{2}(g)} + \mathrm{O_{2}(g)} \rightarrow 2\mathrm{H_{2}O(l)}2H2(g)+O2(g)→2H2O(l)
Relative atomic masses: H = 1, O = 16
The reaction stops while some hydrogen remains in the reaction vessel. State the limiting reactant.
Explain what the unreacted hydrogen shows about why the formation of water has stopped.
A separate reaction uses 0.64 g of oxygen and more hydrogen than is needed. Calculate the mass of water formed.
Five mixtures are prepared. Each contains the same initial mass of oxygen but a different mass of hydrogen. The mass of water obtained from each mixture is shown in the table.
| Mass of hydrogen / g | Mass of water obtained / g |
|---|---|
| 0.008 | 0.072 |
| 0.016 | 0.144 |
| 0.024 | 0.216 |
| 0.032 | 0.216 |
| 0.040 | 0.216 |
Describe the relationship between the two measured masses in the first three mixtures.
Explain why the final two mixtures both produce 0.216 g of water.
142 exam-style questions on Edexcel GCSE Chemistry 2.6 Calculations involving masses, covering 2.6.1 Relative formula mass and percentage composition, 2.6.2 Empirical and molecular formulae, 2.6.3 Conservation of mass, 2.6.4 Reacting masses and concentration in g dm⁻³, 2.6.5 The mole and the Avogadro constant, and 2.6.6 Limiting reactants and stoichiometry. Each one has a worked solution and a mark scheme showing where the marks go.