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Question 23

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)
a.

0.3924 g0.3924\text{ g}0.3924 g of zinc was added to 25.0 cm325.0\text{ cm}^325.0 cm3 of 0.400 mol/dm30.400\text{ mol/dm}^30.400 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.

[1]
b.

Calculate the amount, in moles, of HCl\text{HCl}HCl in the 25.0 cm325.0\text{ cm}^325.0 cm3 of hydrochloric acid.

[1]
c.

Use your answers from (a) to determine which of the reactants is in excess. Show your reasoning.

[3]

Chemical formulae, equations and calculations Questions

  1. IGCSE
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  3. /Chemical formulae, equations and calculations