Question 3.81: If 100.0 g of dinitrogen tetroxide reacts with 100.0 g of hy......
If 100.0 g of dinitrogen tetroxide reacts with 100.0 g of hydrazine ( \text{N}_2\text{H}_4) , what is the theoretical yield of nitrogen if no side reaction takes place? First, we need to identify the limiting reactant. The limiting reactant can be used to calculate the theoretical yield . Determine the amount of limiting reactant required to produce 10.0 grams of NO . Reduce the amount of limiting reactant by the amount used to produce NO. The reduced amount of limiting reactant is then used to calculate an “actual yield.” The “actual” and theoretical yields will give the maximum percent yield.
Determining the limiting reactant :
\text{N}_2\text{ from N}_2\text{O}_4=(100.0\text{ g N}_2\text{O}_4)\left\lgroup\frac{1\text{ mol N}_2\text{O}_4}{92.02\text{ g N}_2\text{O}_4} \right\rgroup \left\lgroup\frac{3\text{ mol N}_2}{1\text{ mol N}_2\text{O}_4} \right\rgroup =3.26016\text{ mol N}_2 \\ \text{N}_2\text{ from N}_2\text{H}_4=(100.0\text{ g N}_2\text{H}_4)\left\lgroup\frac{1\text{ mol N}_2\text{H}_4}{32.05\text{ g N}_2\text{H}_4} \right\rgroup \left\lgroup\frac{3\text{ mol N}_2}{2\text{ mol N}_2\text{H}_4} \right\rgroup =4.68019\text{ mol N}_2\text{N}_2\text{O}_4 is the limiting reactant.
How much limiting reactant used to produce 100.0 g NO?
Determine the “actual yield . “
Amount of \text{N}_2\text{O}_4 available to produce \text{N}_2 = 100.0\text{ g N}_2\text{O}_4 – mass of \text{N}_2\text{O}_4 required to produce 10.0 g NO
100.0\text{ g – 10.221 g = 89.779 g N}_2\text{O}_4 ( unrounded)
Theoretical yield = [“Actual yield” / theoretical yield] x 100%
[(82.01285 \text{ g N}_2) / (91.3497 \text{ g N}_2)] x 100% = 89.7790 = 89.8%