Question

Determine the limiting reagent (lr) and the mass (in g) of nitrogen that can be formed from 50.0g N₂O₄ and 45.0 g N₂H₄. some possibly useful molar masses are as follows: N₂O₄ = 92.02g/mol, N₂H₄=32.05g/mol.

Answer 1

The limiting reagent is N₂O₄, and the mass of nitrogen that can be formed is 23.76 g.

Step-by-step explanation:

The limiting reagent is the reactant that is completely consumed in a chemical reaction, thereby limiting the amount of product that can be formed. To determine the limiting reagent, we need to compare the number of moles of each reactant present and use stoichiometry to calculate the number of moles of product that can be formed.

First, let's convert the mass of each reactant to moles:

1. For N₂O₄: 50.0 g / 92.02 g/mol = 0.543 mol
2. For N₂H₄: 45.0 g / 32.05 g/mol = 1.404 mol

Next, let's use the balanced chemical equation to determine the ratio of moles between the reactants and the product:

2 N₂O₄ + N₂H₄ -> 3 N₂ + 4 H₂O

According to the equation, 2 moles of N₂O₄ react with 1 mole of N₂H₄ to produce 3 moles of N₂.

To determine the limiting reagent, we compare the mole ratios of the reactants to the stoichiometry of the equation. Since N₂H₄ has a larger number of moles (1.404 mol) compared to N₂O₄ (0.543 mol), N₂O₄ is the limiting reagent.

To calculate the mass of nitrogen formed, we use the mole ratio between N₂O₄ and N₂:

0.543 mol N₂O₄ * (3 mol N₂ / 2 mol N₂O₄) * (28.02 g/mol) = 23.76 g N₂