You are given the following decomposition reaction of ammonia
NH3 ---> H2 + N3 unbalanced
When you decompose 12.0g of NH3you produce 1.87 g of H2 Calculate the percent yield of H2
I believe you goofed on the formula (in addition to not balancing it).
2NH3 ==> 3H2 + N2
mols12.0g NH3 = grams/molar mass = ?
Convert mols NH3 to mols H2 using the coefficients in the balanced equation.
Convert mols H2 to grams. g = mols x molasr mass. This is the theoretical yield. The actual yield according to the problem is 1.87g.
%yield = (actual yield/theor yield)*100 = ?
87.6%
To calculate the percent yield of H2, you need to compare the actual yield (1.87 g) with the theoretical yield. The theoretical yield is the maximum amount of H2 that could be produced based on the balanced chemical equation.
First, let's calculate the molar mass of NH3, H2, and N2:
- NH3 (ammonia): Atomic mass of N (14.01 g/mol) + 3 x atomic mass of H (1.01 g/mol) = 17.04 g/mol
- H2 (hydrogen gas): 2 x atomic mass of H = 2.02 g/mol
Now, let's convert the mass of NH3 and H2 to moles:
- Moles of NH3: 12.0 g / 17.04 g/mol = 0.704 mol
- Moles of H2: 1.87 g / 2.02 g/mol = 0.924 mol
We can use the balanced chemical equation to determine the mole ratio between NH3 and H2. From the equation NH3 → H2 + N2, we can see that the mole ratio is 1:1. So, 0.704 mol of NH3 should produce 0.704 mol of H2.
Now, let's calculate the theoretical yield of H2 (the maximum amount of H2 that can be produced):
Theoretical yield = 0.704 mol x (2.02 g/mol) = 1.42 g
Finally, we can calculate the percent yield of H2:
Percent yield = (Actual yield / Theoretical yield) x 100
Percent yield = (1.87 g / 1.42 g) x 100 = 131.7%
Therefore, the percent yield of H2 is 131.7%.
To calculate the percent yield of H2 in this decomposition reaction, you need to compare the actual yield (1.87 g) to the theoretical yield. The theoretical yield is the amount of H2 that would be obtained if the reaction proceeded perfectly, assuming complete conversion of NH3 to H2.
First, let's calculate the theoretical yield of H2.
1. Calculate the molar mass of NH3:
Molar mass of NH3 = (1 × Molar mass of N) + (3 × Molar mass of H)
= (1 × 14.01 g/mol) + (3 × 1.01 g/mol)
= 14.01 g/mol + 3.03 g/mol
= 17.04 g/mol
2. Convert the mass of NH3 (12.0 g) to moles:
Moles of NH3 = mass of NH3 / molar mass of NH3
= 12.0 g / 17.04 g/mol
≈ 0.704 mol
3. Use the balanced equation to determine the stoichiometry between NH3 and H2:
NH3 ---> H2 + N2
From the balanced equation, we can see that 1 mole of NH3 produces 1 mole of H2.
Therefore, moles of H2 = moles of NH3 = 0.704 mol
4. Calculate the molar mass of H2:
Molar mass of H2 = 2 × Molar mass of H
= 2 × 1.01 g/mol
= 2.02 g/mol
5. Calculate the theoretical yield of H2:
Theoretical yield of H2 = moles of H2 × molar mass of H2
= 0.704 mol × 2.02 g/mol
≈ 1.42 g
Now we can calculate the percent yield:
Percent yield = (actual yield / theoretical yield) × 100%
Percent yield = (1.87 g / 1.42 g) × 100%
= 131.7%
Therefore, the percent yield of H2 in this reaction is approximately 131.7%.