Question

2N2H4 + N2O4 → 3N2 + 4H2O

Answer 1

Answer : The amount of $N_2H_4$ will be, 64 g

Solution : Given,

Mass of $N_2O_4$ = 92 g

Molar mass of $N_2O_4$ = 92 g/mole

Molar mass of $N_2H_4$ = 32 g/mole

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

$\text{Moles of }N_2O_4=\frac{\text{Mass of }N_2O_4}{\text{Molar mass of }N_2O_4}=\frac{92g}{92g/mole}=1mole$

Now we have to calculate the moles of $N_2H_4$.

The balanced chemical reaction is,

$2N_2H_4+N_2O_4\rightarrow 3N_2+4H_2O$

From the balanced reaction, we conclude that

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

and from the calculated moles,

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

Now we have to calculate the mass of $N_2H_4$

$\text{Mass of }N_2H_4=\text{Moles of }N_2H_4\times \text{Molar mass of }N_2H_4$

$\text{Mass of }N_2H_4=(2mole)\times (32g/mole)=64g$

Therefore, the amount of $N_2H_4$ will be, 64 g

Answer 2

With N2O4's molar mass of 92 grams/mole, there is 1 mole of N2O4 being used.
For every one mole of N2O4 used, two moles of  N2H4 are used.
2 moles of N2H4 times the molar mass of N2H4 (32g/mol)
2 moles * (32 grams/mole) = 64 grams