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Iron bromide. Iron bromide (FeBr3)
Answer:
%yield of NH₃ = 30%
Explanation:
Actual yield of NH₃ = 40.8g
Theoretical yield = ?
Equation of reaction
N₂ + 3H₂ → 2NH₃
Molar mass of NH₃ = 17g/mol
Molarmass of N = 14.00
2 molecules of N = 2 * 14.00 = 28g/mol
Number of moles = mass / molar mass
Mass = number of moles * molar mass
Mass = 1 * 28.00 = 28g of N₂ (the number of moles of N₂ from the equation is 1).
From the equation of reaction,
28g of N₂ produce (2 * 17)g of NH₃
28g of N₂ = 34g of NH₃
112g of N₂ = x g of NH₃
X = (112 * 34) / 28
X = 136g of NH₃
Theoretical yield = 136g of NH₃
% yield = (actual yield / theoretical yield) * 100
% yield = (40.8 / 136) * 100
% yield = 0.3 * 100
% yield = 30%
