Final answer:
To calculate the percent yield of ammonia, we compare the actual yield of 2.03 g to the theoretical yield of 2.4318 g, resulting from the reaction of nitrogen and hydrogen. The percent yield is calculated as (Actual Yield / Theoretical Yield) * 100%, giving approximately 83.5%, which doesn't match the answer choices.
Step-by-step explanation:
The student's question is about calculating the percent yield of ammonia in a chemical reaction between nitrogen and hydrogen gases. To determine the percent yield, one must compare the actual mass of ammonia collected to the theoretical yield, which is the mass of ammonia expected from stoichiometry, assuming complete reaction. The balanced chemical equation for this reaction is:
N2(g) + 3H2(g) → 2NH3(g)
To calculate the theoretical yield, we need to identify the limiting reactant. Given that nitrogen has a molar mass of 28.02 g/mol and hydrogen has a molar mass of 2.02 g/mol, we can calculate the moles of each reactant:
- Moles of N2 = 2.00 g / 28.02 g/mol = 0.0714 mol
- Moles of H2 = 3.00 g / 2.02 g/mol = 1.4851 mol
Using the molar ratios from the balanced equation (1 mol N2:3 mol H2), nitrogen is the limiting reactant.
The theoretical yield of ammonia would be twice the moles of N2, since the stoichiometry is 1 N2 : 2 NH3. Thus, the theoretical yield in moles of ammonia is 2 * 0.0714 mol = 0.1428 mol. Since the molar mass of ammonia (NH3) is 17.03 g/mol, the theoretical yield in grams is 0.1428 mol * 17.03 g/mol = 2.4318 g.
To calculate the percent yield, use the formula:
Percent Yield = (Actual Yield / Theoretical Yield) * 100%
For this experiment:
- Actual Yield = 2.03 g
- Theoretical Yield = 2.4318 g
- Percent Yield = (2.03 g / 2.4318 g) * 100% ≈ 83.5%
However, none of the answer choices match this value, which indicates that there might have been an error in the calculation or the provided answer choices.