Question

Can someone please help?

Answer 1

Answer: The correct option is, (a) 9.0 g

Explanation : Given,

Mass of $N_2H_4$ = 8.0 g

Mass of $N_2O_4$ = 92 g

Molar mass of $N_2H_4$ = 32 g/mol

Molar mass of $N_2O_4$ = 92 g/mol

First we have to calculate the moles of $N_2H_4$ and $N_2O_4$.

$\text{Moles of }N_2H_4=\frac{\text{Given mass }N_2H_4}{\text{Molar mass }N_2H_4}$

$\text{Moles of }N_2H_4=\frac{8.0g}{32g/mol}=0.25mol$

and,

$\text{Moles of }N_2O_4=\frac{\text{Given mass }N_2O_4}{\text{Molar mass }N_2O_4}$

$\text{Moles of }N_2O_4=\frac{92g}{92g/mol}=1mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

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

From the balanced reaction we conclude that

As, 2 mole of $N_2H_4$ react with 1 mole of $N_2O_4$

So, 0.25 moles of $N_2H_4$ react with $\frac{0.25}{2}=0.125$ moles of $N_2O_4$

From this we conclude that, $N_2O_4$ is an excess reagent because the given moles are greater than the required moles and $N_2H_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 2 mole of $N_2H_4$ react to give 4 mole of $H_2O$

So, 0.25 mole of $N_2H_4$ react to give $\frac{4}{2}\times 0.25=0.5$ mole of $H_2O$

Now we have to calculate the mass of $H_2O$

$\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O$

Molar mass of $H_2O$ = 18 g/mole

$\text{ Mass of }H_2O=(0.5moles)\times (18g/mole)=9.0g$

Therefore, the mass of $H_2O$ produced is, 9.0 grams.