## Chemistry: Molecular Approach (4th Edition)

0.5 mol of $O_2$ in the mixture.
1. Identify the limiting reactant. Find the amount of product if each reactant is completely consumed. $$4.2 \space moles \space ZnS \times \frac{ 2 \space moles \ ZnO }{ 2 \space moles \space ZnS } = 4.2 \space moles \space ZnO$$ $$6.8 \space moles \space O_2 \times \frac{ 2 \space moles \ ZnO }{ 3 \space moles \space O_2 } = 4.5 \space moles \space ZnO$$ Since the reaction of $ZnS$ produces less $ZnO$ for these quantities, it is the limiting reactant. 2. Find the amount of $O_2$ consumed: $$4.2 \space moles \space ZnS \times \frac{3 \space moles \space O_2}{2 \space moles \space ZnS} = 6.3 \space moles \space O_2$$ 3. Subtract that amount from the amount in the reaction mixture. $$Excess = 6.8 - 6.3 = 0.5 \space mol \space O_2$$