Question
What is the theoretical yield in grams for this reaction under the given conditions?
1.52 g H2 is allowed to react with 10.1 g N2, producing 1.26 g NH3.
1.52 g H2 is allowed to react with 10.1 g N2, producing 1.26 g NH3.
Answers
GPT 3.5
To find the theoretical yield of NH3, we need to calculate the limiting reactant in the reaction between H2 and N2.
First, we need to determine the moles of each reactant:
- Moles of H2: 1.52 g / 2.02 g/mol = 0.753 moles
- Moles of N2: 10.1 g / 28.02 g/mol = 0.360 moles
Next, we need to determine the mole ratio between H2 and N2 in the balanced chemical equation:
H2 + N2 -> 2NH3
The ratio is 1:1, which means that H2 is the limiting reactant.
Now, we need to calculate the theoretical yield of NH3 using the moles of H2:
- Moles of NH3 produced = Moles of H2 used = 0.753 moles
- Mass of NH3 produced = Moles of NH3 produced x molar mass of NH3
= 0.753 moles x 17.03 g/mol = 12.81 g
Therefore, the theoretical yield of NH3 in grams is 12.81 g.
First, we need to determine the moles of each reactant:
- Moles of H2: 1.52 g / 2.02 g/mol = 0.753 moles
- Moles of N2: 10.1 g / 28.02 g/mol = 0.360 moles
Next, we need to determine the mole ratio between H2 and N2 in the balanced chemical equation:
H2 + N2 -> 2NH3
The ratio is 1:1, which means that H2 is the limiting reactant.
Now, we need to calculate the theoretical yield of NH3 using the moles of H2:
- Moles of NH3 produced = Moles of H2 used = 0.753 moles
- Mass of NH3 produced = Moles of NH3 produced x molar mass of NH3
= 0.753 moles x 17.03 g/mol = 12.81 g
Therefore, the theoretical yield of NH3 in grams is 12.81 g.