You began with 22 grams of nitrogen, and produced 19 grams of ammonia, #NH_3#, According to the the mass calculations, you should be able to produce 26 grams of ammonia. What is the percent yield? #N_2 + 3H_2 ->2NH_3#

73% of the yield is produced.

The percent yield formula is

The percent yield is calculated by dividing the actual yield (19 grams) by the theoretical yield (26 grams), then multiplying by 100.

Percent Yield = (Actual Yield / Theoretical Yield) * 100

Percent Yield = (19 grams / 26 grams) * 100 ≈ 73.1%

To calculate the percent yield, you first need to determine the theoretical yield, which is the maximum amount of product that can be produced based on stoichiometry. Then, you use the actual yield (the amount of product obtained in the experiment) to calculate the percent yield using the formula:

[ \text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100% ]

Given that you began with 22 grams of nitrogen and produced 19 grams of ammonia, you can use stoichiometry to find the theoretical yield of ammonia based on the balanced chemical equation:

[ N_2 + 3H_2 \rightarrow 2NH_3 ]

1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. Since the molar mass of nitrogen is 14 grams/mole and the molar mass of ammonia is 17 grams/mole, we can calculate the theoretical yield:

[ \text{Moles of nitrogen} = \frac{22 , \text{grams}}{14 , \text{grams/mole}} = 1.57 , \text{moles} ] [ \text{Theoretical moles of ammonia} = 2 \times 1.57 = 3.14 , \text{moles} ] [ \text{Theoretical yield of ammonia} = 3.14 , \text{moles} \times 17 , \text{grams/mole} = 53.38 , \text{grams} ]

Now, we can calculate the percent yield:

[ \text{Percent yield} = \frac{19 , \text{grams}}{53.38 , \text{grams}} \times 100% \approx 35.6% ]

Therefore, the percent yield of the reaction is approximately 35.6%.