QUESTION 11
Ammonia gas is formed from nitrogen gas and hydrogen gas, according to the following equation: N2 (g) + 3H2 (g) Imported Asset 2NH3 (g). If 84.0 g of nitrogen gas is allowed to react with an excess of hydrogen gas to produce 85.0 g of ammonia, what is the percent yield of this reaction?
42.2%
65.0%
70.3%
83.3%
83.3
To find the percent yield of a reaction, you need to compare the actual yield (the amount of ammonia produced) to the theoretical yield (the amount of ammonia that would be produced if the reaction went to completion).
First, let's determine the theoretical yield of ammonia.
From the balanced chemical equation, you can see that 1 mole of nitrogen gas (N2) reacts with 3 moles of hydrogen gas (H2) to produce 2 moles of ammonia (NH3).
Next, we need to convert the given mass of nitrogen gas to moles.
Using the molar mass of nitrogen gas (N2), which is 28.02 g/mol, we can calculate the number of moles of nitrogen gas:
moles of N2 = mass of N2 / molar mass of N2
= 84.0 g / 28.02 g/mol
= 3.0 mol
Since the ratio from the balanced equation is 1:2, this means that 3.0 mol of nitrogen gas would react to produce twice as many moles of ammonia (NH3):
moles of NH3 (theoretical yield) = 3.0 mol of N2 * (2 mol of NH3 / 1 mol of N2)
= 6.0 mol
Now, let's calculate the actual yield of ammonia:
actual yield = 85.0 g (given)
Finally, we can calculate the percent yield using the formula:
percent yield = (actual yield / theoretical yield) * 100
percent yield = (85.0 g / 6.0 mol) * 100
percent yield ≈ 65.0%
Therefore, the percent yield of this reaction is approximately 65.0%.