What is the theoretical yield in grams for this reaction under the given conditions? 1.31 g H2 is allowed to react with 9.95 g N2, producing 1.12 g NH3.
First, we need to determine the limiting reagent in the reaction.
1. Convert the given masses of H2 and N2 into moles using their molar masses:
- H2: 1.31 g / 2.016 g/mol = 0.650 mol
- N2: 9.95 g / 28.02 g/mol = 0.355 mol
2. Calculate the mole ratio between H2 and NH3 in the balanced chemical equation:
3H2 + N2 → 2NH3
1 mol H2 produces 2 mol NH3
3. Determine the maximum amount of NH3 that can be produced by each reactant:
- From H2: 0.650 mol H2 × 2 mol NH3/mol H2 = 1.30 mol NH3
- From N2: 0.355 mol N2 × 2 mol NH3/mol N2 = 0.710 mol NH3
Since N2 limits the amount of NH3 that can be produced, the theoretical yield is 0.710 mol NH3.
4. Calculate the theoretical yield in grams using the molar mass of NH3:
0.710 mol NH3 × 17.03 g/mol = 12.06 g NH3
Therefore, the theoretical yield in grams for this reaction is 12.06 g NH3.