We are asked to calculate the amount (in moles) of the excess reactant.

2 ZnS(s) + 3 O_{2}(g) → 2 ZnO(s) + 2 SO_{2}(g)

Calculate the mole ZnO based on both reactants:

$\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{ZnS}\mathbf{}}\mathbf{\times}\mathbf{}\frac{\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{\hspace{0.17em}}\mathbf{ZnO}}{\mathbf{2}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{ZnS}\mathbf{}}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{ZnO}\phantom{\rule{0ex}{0ex}}\mathbf{8}\mathbf{.}\mathbf{7}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{O}}_{\mathbf{2}}\mathbf{}}\mathbf{\times}\mathbf{}\frac{\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{\hspace{0.17em}}\mathbf{ZnO}}{\mathbf{3}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{O}}_{\mathbf{2}}\mathbf{}}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{ZnO}\phantom{\rule{0ex}{0ex}}$

Since ZnS produced less product, it is limiting.

Calculate O_{2} used.

$\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{ZnS}\mathbf{}}\mathbf{\times}\mathbf{}\frac{\mathbf{3}\mathbf{}\mathbf{mol}\mathbf{\hspace{0.17em}}{\mathbf{O}}_{\mathbf{2}}}{\mathbf{2}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{ZnS}\mathbf{}}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{O}}_{\mathbf{2}}$

Zinc sulfide reacts with oxygen according to the reaction:

2 ZnS(s) + 3 O_{2}(g) → 2 ZnO(s) + 2 SO_{2}(g)

A reaction mixture initially contains 4.2 mol ZnS and 8.7 mol O_{2}.

Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant remains?

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