Zinc reacts with dilute hydrochloric acid. The equation for the reaction is:
Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) \text{Zn(s)} + 2\text{HCl(aq)} \rightarrow \text{ZnCl}_2\text{(aq)} + \text{H}_2\text{(g)} Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)0.327 g0.327\text{ g}0.327 g of zinc was added to 40.0 cm340.0\text{ cm}^340.0 cm3 of 0.220 mol/dm30.220\text{ mol/dm}^30.220 mol/dm3 hydrochloric acid. (Assume the relative atomic mass of zinc, Ar(Zn)=65.4A_r(\text{Zn}) = 65.4Ar(Zn)=65.4). (i) Calculate the amount, in moles, of zinc used. (ii) Calculate the amount, in moles, of HCl\text{HCl}HCl in the 40.0 cm340.0\text{ cm}^340.0 cm3 of hydrochloric acid.
Use your answers from (a) to determine which of the reactants is in excess. Show your reasoning.
