Verified Answer :

premedchemist

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

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

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