If liquid mercury and oxygen combine to make mercury(II) oxide, what is the percent yield of HgO if 350.0 grams of Hg( l ) react with 150.0 g of O₂( g ) to produce 272.3 grams of the product?

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Quizplus · Accepted Answer

The percent yield is calculated using the formula: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$. First, determine the limiting reactant. The molar mass of Hg is approximately 200.59 g/mol, and for O2, it is approximately 32.00 g/mol. The reaction is $2Hg + O_2 \rightarrow 2HgO$. For Hg, $350.0 \, \text{g} \times \frac{1 \, \text{mol Hg}}{200.59 \, \text{g}} = 1.744 \, \text{mol Hg}$. For O2, $150.0 \, \text{g} \times \frac{1 \, \text{mol O}_2}{32.00 \, \text{g}} = 4.688 \, \text{mol O}_2$. Hg is the limiting reactant. The theoretical yield of HgO from the limiting reactant is $1.744 \, \text{mol Hg} \times \frac{2 \, \text{mol HgO}}{2 \, \text{mol Hg}} \times \frac{216.59 \, \text{g HgO}}{1 \, \text{mol HgO}} = 377.6 \, \text{g HgO}$. The percent yield is $\frac{272.3 \, \text{g}}{377.6 \, \text{g}} \times 100\% = 72.07\%$.

If Liquid Mercury and Oxygen Combine to Make Mercury(II) Oxide