Aluminum reacts with dilute hydrochloric acid to form aluminum chloride and hydrogen gas:
2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g) 2\text{Al(s)} + 6\text{HCl(aq)} \rightarrow 2\text{AlCl}_3\text{(aq)} + 3\text{H}_2\text{(g)} 2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)A 0.162 g0.162\text{ g}0.162 g sample of pure aluminum is added to 45.0 cm345.0\text{ cm}^345.0 cm3 of 0.500 mol dm−30.500\text{ mol dm}^{-3}0.500 mol dm−3 hydrochloric acid.
One of these reagents is in excess and the other reagent limits the amount of hydrogen produced in the reaction.
Calculate the maximum volume, in m3\text{m}^3m3, of hydrogen gas produced at 60 ∘C60\,^\circ\text{C}60∘C and 96.0 kPa96.0\text{ kPa}96.0 kPa.
Give your answer to 3 significant figures.
In your answer you should identify the limiting reagent in the reaction.
The gas constant, R=8.31 J K−1 mol−1R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}R=8.31 J K−1 mol−1
Molar mass of Al=27.0 g mol−1\text{Al} = 27.0\text{ g mol}^{-1}Al=27.0 g mol−1
79 exam-style questions on AQA A Level Chemistry Amount of substance, covering Relative atomic mass and relative molecular mass, The mole and the Avogadro constant, The ideal gas equation, Empirical and molecular formula, and Balanced equations and associated calculations. Each one has a worked solution and a mark scheme showing where the marks go.