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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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Limiting Reagent concept. If you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems.