Answer to Question #147561 in Inorganic Chemistry for Cherry Delos reyes

Question #147561
What is the % yield of 7.50g of nitrogen reacted with 1.30g of hydrogen and 5.75 g of ammonia is produced
1
Expert's answer
2020-12-01T09:05:30-0500

N2 + 3H2 = 2NH3

n(N2) = m(N2)/Mr(N2) = m(N2)/(2*Ar(N)) = (7.50 g)/(2*14 g/mol) = 0.267857 mol

n(H2) = m(H2)/Mr(H2) = m(H2)/(2*Ar(H)) = (1.30 g)/(2*1 g/mol) = 0.65 mol

1/3*n(H2) = 0.216667 mol < n(N2) so we have the excess of nitrogen and we will calculate yield from n(H2)

n(NH3) theoretical = 2/3*n(H2) = 0.433333 mol;

m(NH3) theoretical = n(NH3)*Mr(NH3) = n(NH3)*(Ar(N)+3*Ar(H)) = (0.433333 mol)*(14 g/mol + 3*1 g/mol) = 7.367 g

yield = m(NH3)produced/m(NH3)theoretical = (5.75g/7.367g)*100% = 78.05%


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