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?
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%.