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If 15.0g N2 and 2.00g H2 react to produce 1.38g NH3, what is the percent yield of the reaction?

N2(g)+3H2(g) ---> 2NH3(g)

User Hirt
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1 Answer

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Final answer:

To determine the percent yield, we first calculated the theoretical yield based on the stoichiometry of the reaction and the given amounts of reactants, which was 18.12g of NH3. Then we used the actual yield of 1.38g NH3 to calculate the percent yield as 7.62%.

Step-by-step explanation:

To calculate the percent yield of the reaction N2(g) + 3H2(g) → 2NH3(g), we need to compare the actual yield to the theoretical yield. Firstly, we calculate the theoretical yield of NH3 using stoichiometry:

From the balanced equation, 28 grams of N2 (1 mole) react with 6 grams of H2 (3 moles) to produce 34 grams of NH3 (2 moles).

To find the limiting reactant, calculate the moles of each reactant:

  • For N2: 15.0g × (1 mole / 28g) = 0.536 moles N2
  • For H2: 2.00g × (1 mole / 2g) = 1.00 mole H2

N2 is the limiting reactant, as it will run out first. Therefore, the theoretical yield would be based on the moles of N2:

0.536 moles N2 × (2 moles NH3 / 1 mole N2) × (17g NH3 / 1 mole NH3) = 18.12g NH3

The percent yield is calculated using the actual yield (1.38g) divided by the theoretical yield (18.12g), multiplied by 100%:

Percent Yield = (Actual Yield / Theoretical Yield) × 100%

Percent Yield = (1.38g / 18.12g) × 100%

Percent Yield = 7.62%

User Jeff Chen
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