N2 + 3H2 Right arrow. 2NH3 What is the percent yield of NH3 if the reaction of 26.3 g of H2 produces 79.0 g of NH3? Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100..
2 answers:
Answer:
53.4%
Explanation:
N₂ + 3H₂ → 2NH₃
Given that:
26.3 g of H₂ react with N₂ to produce 79.0 g of NH₃
Then:
The number of moles of H₂ = 26.3g of H₂ * (1 mol of H₂/ 2.02g of H₂)
= 13.05 mol of H₂
The number of moles of NH₃ = 13.05 mol of H₂ * ( 2mol of NH₃/ 3 mol of H₂)
= 8.697 mol of NH₃
The mass of NH₃ = 8.697 mol of NH₃ *( 17.04g of NH₃/ 1 mol of NH₃) = 148.1 g of NH₃
The percent yield = actual yield/ theoretical yield * 100%
The percent yield = ( 79.0 g/ 148.1 g )* 100%
The percent yield ≅ 53.4 %
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