Question

Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N204 and 45.0 g N2H4. Some possibly useful molar masses are as follows: N2O4 92.02 g/mol, N2H4 32.05 g/mol N204) 2 N2H4(1)3 N2(g) + 4 H2O(g)

Answer 1

Answer: The mass of nitrogen gas produced will be 45.64 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$    .....(1)

• For $N_2O_4$

Given mass of $N_2O_4=50.0g$

Molar mass of $N_2O_4=92.02g/mol$

Putting values in equation 1, we get:

$\text{Moles of }N_2O_4=\frac{50g}{92.02g/mol}=0.543mol$

• For $N_2H_4$

Given mass of $N_2H_4=45.0g$

Molar mass of $N_2O_4=32.05g/mol$

Putting values in  equation 1, we get:

$\text{Moles of }N_2H_4=\frac{45g}{32.05g/mol}=1.40mol$

For the given chemical reaction:

$N_2O_4(l)+2N_2H_4(l)\rightarrow 3N_2(g)+4H_2O(g)$

By stoichiometry of the reaction:

1 mole of $N_2O_4$ reacts with 2 moles of $N_2H_4$

So, 0.543 moles of $N_2O_4$ will react with = $\frac{2}{1}\times 0.543=1.086moles$ of $N_2H_4$

As, the given amount of $N_2H_4$ is more than the required amount. Thus, it is considered as an excess reagent.

Hence, $N_2O_4$ is the limiting reagent.

By Stoichiometry of the reaction:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 0.543 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 0.543=1.629moles$ of nitrogen gas.

Now, calculating the mass of nitrogen gas from equation 1, we get:

Molar mass of nitrogen gas = 28.02 g/mol

Moles of nitrogen gas = 1.629 moles

Putting values in equation 1, we get:

$1.629mol=\frac{\text{Mass of nitrogen gas}}{28.02g/mol}\\\\\text{Mass of nitrogen gas}=45.64g$

Hence, the mass of nitrogen gas produced will be 45.64 grams.