Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N₂O₄ and 45.0 g N₂H₄. Some possibly useful molar masses are as follows: N₂O₄ = 92.02 g mol^-1, N₂H₄ = 32.05 g mol^-1. N₂O₄(l) + 2N₂H₄(l) → 3N₂(g) + 4H₂O(g) A) LR = N₂H₄, 59.0 g N₂formed B) LR = N₂O₄, 105 g N₂formed C) LR = N₂O₄, 45.7 g N₂formed D) LR = N₂H₄, 13.3 g N₂formed E) No LR, 45.0 g N₂formed

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The balanced equation is:N₂O₄(l) + 2N₂H₄(l) → 3N₂(g) + 4H₂O(g)Moles of N₂O₄ = 50.0 g / 92.02 g/mol = 0.543 molMoles of N₂H₄ = 45.0 g / 32.05 g/mol = 1.40 molThe mole ratio is 1:2, so N₂O₄ is the limiting reactant.Moles of N₂ formed = 0.543 mol N₂O₄ x (3 mol N₂ / 1 mol N₂O₄) = 1.63 mol N₂Mass of N₂ formed = 1.63 mol N₂ x 28.02 g/mol = 45.7 g N₂

Determine the Limiting Reactant (LR) and the Mass (In G)