Question

1.30 grams of H₂ are reacted with an excess of N₂ to produce 4.21 grams of NH3-

3H₂ + N2 → 2NH3
What was the percent yield for ammonia in this reaction?

A-27.40%
B-28.80%
C-57.50%
D-62.60%

Answer 1

To find the percent yield, we first need to calculate the theoretical yield, which is the amount of NH3 that would be produced if the reaction went to completion. We can do this by calculating the amount of NH3 that would be produced from the amount of H2 used in the reaction:

From the balanced chemical equation, we see that 3 moles of H2 react with 1 mole of N2 to produce 2 moles of NH3. Therefore, the number of moles of NH3 produced is:

moles of NH3 = (1.30 g H2) / (2.016 g/mol H2) x (2 mol NH3 / 3 mol H2) = 0.427 mol NH3

The molar mass of NH3 is 17.031 g/mol, so the theoretical yield of NH3 is:

theoretical yield = (0.427 mol NH3) x (17.031 g/mol NH3) = 7.28 g NH3

Now we can calculate the percent yield:

percent yield = (actual yield / theoretical yield) x 100%

The actual yield is given as 4.21 g NH3, so:

percent yield = (4.21 g NH3 / 7.28 g NH3) x 100% ≈ 57.85%

Therefore, the percent yield for ammonia in this reaction is approximately 57.85%, which is closest to option C (57.50%).