What is the limiting reagent when 150.0g of nitrogen react with 32.1g of hydrogen
N2(g)+ 3H2(g) -> 2NH3(g)
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Expert's answer
2016-02-16T00:00:57-0500
m(N2) = 150.0 g; m(H2) = 32.1 g; The molar mass of nitrogen is M(N2) = 28.01 g/mol; The amount of nitrogen is n(N2) = m(N2)/M(N2) = 150.0/28.01 = 5.35 moles; The molar mass of hydrogen is M(H2) = 2.016 g/mol; The amount of hydrogen is n(H2) = m(H2)/M(H2) =32.1/2.016 = 15.92 moles; The reaction stoichiometry says that 1 mole of nitrogen requires 3 moles of hydrogen, so for 5.35 moles of nitrogen 5.35×3 = 16.05 moles of hydrogen required. But only 15.92 moles of hydrogen are present, therefore, the hydrogen is the limiting reactant. Answer: hydrogen is the limiting reactant.
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