1.86 g h2 is allowed to react with 9.70 g n2, producing 1.24 g nh3. part a what is the theoretical yield in grams for this reaction under the given conditions?
What is the theoretical yield in grams for this reaction under the given conditions?
Answer:
In order to determine the theoretical yield of a chemical reaction, we need to first identify the limiting reactant. The limiting reactant is the reactant that will be completely consumed in the reaction, thus limiting the amount of product that can be formed.
- Calculate the number of moles of each reactant:
For H2:
n_{H_2} = \frac{1.86 g}{2.016 g/mol} = 0.9226 mol
For N2:
n_{N_2} = \frac{9.70 g}{28.02 g/mol} = 0.3460 mol
- Determine the stoichiometry of the reaction:
The balanced chemical equation for the reaction is:
3H_2 + N_2 \rightarrow 2NH_3
- Identify the limiting reactant:
Calculate the moles of NH3 that can be produced by each reactant:
- For H2: (0.9226 mol H_2 \times \frac{2 mol NH_3}{3 mol H_2} = 0.6150 mol NH_3)
- For N2: 0.3460 mol N_2 \times \frac{2 mol NH_3}{1 mol N_2} = 0.6920 mol NH_3
H2 is the limiting reactant as it produces fewer moles of NH3.
- Calculate the theoretical yield of NH3:
Using the molar mass of NH3 (17.03 g/mol):
m_{NH_3} = 0.6150 mol NH_3 \times 17.03 g/mol = 10.4695 g
Therefore, the theoretical yield of NH3 under the given conditions is 10.4695 grams.