QUESTION 10

Ammonia gas is formed from nitrogen gas and hydrogen gas according to the following equation: N2 (g) + 3H2 (g) Imported Asset 2NH3 (g). If 112 grams of nitrogen gas is allowed to react with an excess of hydrogen gas to produce 120 grams of ammonia, what is the percent yield of this reaction?

44.1%

66.2%

88.3%

96.4%

44.1%

To calculate the percent yield of a reaction, you need to compare the actual yield to the theoretical yield. The actual yield is the amount of product obtained from the reaction, while the theoretical yield is the amount of product that would be obtained if the reaction proceeded perfectly, based on stoichiometry.

In this case, the 112 grams of nitrogen gas is the limiting reactant, meaning it restricts the amount of product that can be formed. To find the theoretical yield of ammonia, we need to use stoichiometry and convert the mass of nitrogen gas to the mass of ammonia.

According to the balanced equation, 1 mole of nitrogen gas reacts with 2 moles of ammonia. The molar mass of nitrogen gas (N2) is 28 grams/mol, and the molar mass of ammonia (NH3) is 17 grams/mol.

Step 1: Calculate the moles of nitrogen gas (N2):
moles of N2 = mass of N2 / molar mass of N2
moles of N2 = 112g / 28 g/mol
moles of N2 = 4 moles

Step 2: Use the ratio from the balanced equation to determine the moles of ammonia (NH3) that can be formed:
moles of NH3 = moles of N2 x (2 moles of NH3 / 1 mole of N2)
moles of NH3 = 4 moles x (2/1)
moles of NH3 = 8 moles

Step 3: Calculate the theoretical yield of ammonia (NH3) in grams:
mass of NH3 = moles of NH3 x molar mass of NH3
mass of NH3 = 8 moles x 17 g/mol
mass of NH3 = 136 grams

The theoretical yield of ammonia is 136 grams.

Step 4: Calculate the percent yield:
percent yield = (actual yield / theoretical yield) x 100%
percent yield = (120g / 136g) x 100%
percent yield = 0.882 x 100%
percent yield = 88.2%

Therefore, the percent yield of the reaction is approximately 88.3%. Therefore, the closest answer choice is 88.3%.

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