To determine the percent yield, we first need to calculate the theoretical yield of NH3 based on the limiting reactant.
1. Calculate the moles of each reactant:
- Moles of H2 = 1.31g / 2.02g/mol = 0.65 mol
- Moles of N2 = 9.95g / 28.02g/mol = 0.355 mol
2. Determine the limiting reactant:
- In this case, N2 is the limiting reactant since it produces less moles of NH3.
3. Calculate the theoretical yield of NH3 based on the limiting reactant:
- From the balanced equation: 3H2 + N2 -> 2NH3
- Moles of NH3 that can be produced from 0.355 mol N2 = 0.355 mol N2 x (2 mol NH3 / 1 mol N2) = 0.71 mol NH3
- Theoretical yield of NH3 = 0.71 mol x 17.03g/mol = 12.07g NH3
4. Calculate the percent yield:
- Actual yield = 1.12g NH3
- Percent yield = (Actual yield / Theoretical yield) x 100%
- Percent yield = (1.12g / 12.07g) x 100% = 9.29%
Therefore, the percent yield for this reaction under the given conditions is 9.29%.
1.31g H2 is allowed to react with 9.95g N2, producing 1.12 g NH3
What is the percent yield for this reaction under the given conditions?
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