# 22.N2 + 3H2 = 2NH3 1.00 mole of N2 was reacted with 3.00 moles of H2 in the reaction above to produce 28.0g of NH3. What was the % yield of this reaction?122%100%93.7%82.2%

Question

22.

N2 + 3H2 = 2NH3
1.00 mole of N2 was reacted with 3.00 moles of H2 in the reaction above to produce 28.0g of NH3. What was the % yield of this reaction?

122%

100%

93.7%

82.2%

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Step 1

The percentage yield for a reaction is the percent ratio of the actual yield to the theoretical yield of the product formed in the reaction.

The formula to calculate the percentage yield of the reaction is shown on the white board.

In the formula, actual yield is the actual amount of product formed in the reaction, theoretical yield is the amount of product formed from the limiting reactant and it is calculated using the balanced chemical reaction.

The limiting reactant is defined as the reactant that limits the amount of product formed in the reaction. It is present in lesser amount than the other reactant which is present in excess.

Step 2

The given reaction is shown on the white board.

In the balanced chemical reaction, 1 mol of nitrogen gas is reacting with 3 mol of hydrogen gas to produce 2 mol of ammonia gas.

The given number of moles of nitrogen gas and hydrogen gas is 1.00 mol and 3.00 mol respectively. Since, to calculate the theoretical yield, nitrogen and hydrogen gas should be in 1:3 ratio and the given ratio of nitrogen and hydrogen gas is also 1:3 thus, no reactant is present in excess. Therefore, theoretical yield can be calculated from any of the two reactants.

Step 3

Let us calculate the theoretical yield from nitrogen gas.

The given number of moles of nitrogen gas is 1.00 mol. From the balanced chemical reaction, 1 mol of nitrogen gas produces 2 mol of ammonia gas or NH3 gas.

Thus, the number of moles of ammonia gas will be 2 mol.

To calculate the mass of ammonia gas formed, first calculate its molar mass by ...

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