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blsea [12.9K]
3 years ago
9

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..
Chemistry
2 answers:
Romashka [77]3 years ago
7 0

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 %

Andrew [12]3 years ago
3 0

Answer:

53.4

Explanation:

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Explanation:

For the first part,

 Reaction equation:

         N₂  +  3H₂   →   2NH₃

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Unknown:

Number of moles of N₂  = ?

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         2 moles of NH₃ is produced from 1 mole of N₂

        6 moles of NH₃ will be produced from \frac{6}{2}  mole of N₂

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Number of moles of sulfur = 2.4moles

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Mass of sulfur  = ?

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The number of moles of any substance can be found using the expression below;

 Number of moles  = \frac{mass}{molar mass}

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