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,N₂H₄ = 32.05 g/mol. N₂O₄(l)+ 2 N₂H₄(l)→ 3 N₂(g)+ 4 H₂O(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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To find the limiting reactant, we calculate the moles of each reactant and then use the stoichiometry of the reaction. For $N_2O$ (92.02 g/mol), $50.0 \, \text{g} / 92.02 \, \text{g/mol} = 0.543 \, \text{moles}$. For $N_2H_4$ (32.05 g/mol), $45.0 \, \text{g} / 32.05 \, \text{g/mol} = 1.404 \, \text{moles}$. The reaction requires 2 moles of $N_2H_4$ for every mole of $N_2O$, so $N_2O$ is the limiting reactant because the ratio of $N_2H_4$ to $N_2O$ is less than 2:1. From the stoichiometry, 1 mole of $N_2O$ produces 3 moles of $N_2$, so $0.543 \, \text{moles} \times 3 = 1.629 \, \text{moles of } N_2$. The molar mass of $N_2$ is $14.01 \times 2 = 28.02 \, \text{g/mol}$, so $1.629 \, \text{moles} \times 28.02 \, \text{g/mol} = 45.7 \, \text{g}$.

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