1.31g H2 is allowed to react with 9.95g N2, producing 1.12 g NH3
What is the theoretical yield in grams for this reaction under the given conditions?
To find the theoretical yield, we need to first determine the limiting reactant in the reaction. This can be done by calculating the moles of each reactant and comparing them based on the balanced chemical equation.
1. Calculate the moles of each reactant:
- Moles of H2:
n(H2) = 1.31g / 2.02g/mol = 0.65 mol
- Moles of N2:
n(N2) = 9.95g / 28g/mol = 0.36 mol
2. Write and balance the chemical equation for the reaction:
3H2 + N2 -> 2NH3
3. Determine the limiting reactant:
From the balanced chemical equation, we can see that 3 moles of H2 react with 1 mole of N2 to produce 2 moles of NH3.
Let's calculate the moles of NH3 that can be produced from each reactant:
- Moles of NH3 from H2: 0.65 mol H2 x (2 mol NH3 / 3 mol H2) = 0.43 mol NH3
- Moles of NH3 from N2: 0.36 mol N2 x (2 mol NH3 / 1 mol N2) = 0.72 mol NH3
Since the moles of NH3 from N2 is greater than that from H2, N2 is the limiting reactant.
4. Calculate the theoretical yield of NH3 that can be produced using N2:
molar mass of NH3 = 17.03 g/mol
Theoretical yield = 0.36 mol N2 x (2 mol NH3 / 1 mol N2) x 17.03 g/mol = 12.27 g
Therefore, the theoretical yield in grams for this reaction under the given conditions is 12.27 g NH3.