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