Questions: Learning Check In the lab, N2 and 5.0 g of H2 are reacted and produce 16.0 g of NH3. What is the percent yield for the reaction? N2(g) + 3 H2(g) -> 2 NH3(g) 1) 31.3 % 2) 57 % 3) 80.0 %

Solution Steps

First, we need to determine the moles of hydrogen gas (\(\mathrm{H}_2\)) used in the reaction. The molar mass of \(\mathrm{H}_2\) is approximately 2.016 g/mol.

\[ \text{Moles of } \mathrm{H}_2 = \frac{5.0 \, \text{g}}{2.016 \, \text{g/mol}} = 2.480 \, \text{mol} \]

The balanced chemical equation is:

\[ \mathrm{N}_2(\mathrm{~g}) + 3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3(\mathrm{~g}) \]

From the stoichiometry of the reaction, 3 moles of \(\mathrm{H}_2\) react with 1 mole of \(\mathrm{N}_2\). Therefore, the moles of \(\mathrm{N}_2\) required for 2.480 moles of \(\mathrm{H}_2\) are:

\[ \text{Moles of } \mathrm{N}_2 = \frac{2.480 \, \text{mol} \, \mathrm{H}_2}{3} = 0.8267 \, \text{mol} \]

Since the amount of \(\mathrm{N}_2\) is not given, we assume it is in excess, making \(\mathrm{H}_2\) the limiting reactant.

Using the stoichiometry of the reaction, 3 moles of \(\mathrm{H}_2\) produce 2 moles of \(\mathrm{NH}_3\). Therefore, 2.480 moles of \(\mathrm{H}_2\) will produce:

\[ \text{Moles of } \mathrm{NH}_3 = \frac{2.480 \, \text{mol} \, \mathrm{H}_2 \times 2 \, \text{mol} \, \mathrm{NH}_3}{3 \, \text{mol} \, \mathrm{H}_2} = 1.6533 \, \text{mol} \]

The molar mass of \(\mathrm{NH}_3\) is approximately 17.03 g/mol. Therefore, the theoretical yield in grams is:

\[ \text{Theoretical yield} = 1.6533 \, \text{mol} \times 17.03 \, \text{g/mol} = 28.15 \, \text{g} \]

The actual yield of \(\mathrm{NH}_3\) is given as 16.0 g. The percent yield is calculated as:

\[ \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{16.0 \, \text{g}}{28.15 \, \text{g}} \right) \times 100 = 56.83\% \]

Final Answer