Question

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Limiting Reactants and Theoretical Yield Calculations N_2(g) + 3H_2(g) rightarrow 2NH_3(g) If 112 g of N_2 reacts completely with 30.0g H_2. What is the limiting reactant? What is the theoretical yield? How much of the reagent in excess is left over?

Answer 1

For the given reaction,

moles of N2 = 112 g/28 g/mol = 4 mols

moles of H2 = 30.2 g/2 = 15.1 mols

If all of N2 is consumed, we would need = 12 mols

1) Since moles of H2 available is greater than needed, N2 is the limiting reactant

2) theoretical yield of NH3 = 4 mol x 2 x 17 g/mol = 136 g

3) Excess reagent left at the end = (15.1 - 12) mols x 2 = 6.2 g