Question

How many grams of N2 are required to produce 240.0g NH3?

Answer 1

Balanced equation: N2 + 3H2 ==> 2NH3
moles of NH3 in 240.0 g NH3: 240.0 g x 1 mol/17 g = 14.1 moles
moles N2 needed: 14.1 moles NH3 x 1 mole N2/2 moles NH3 = 7.05 moles N2 needed
grams N2 needed: 7.05 moles x 28 g/mole = 197.4 grams needed :)

Answer 2

Answer: The mass of $N_2$ required are, 198 grams.

Explanation : Given,

Mass of $NH_3$ = 240.0  g

Molar mass of $NH_3$ = 17 g/mol

First we have to calculate the moles of $NH_3$.

$\text{Moles of }NH_3=\frac{\text{Given mass }NH_3}{\text{Molar mass }NH_3}$

$\text{Moles of }NH_3=\frac{240.0g}{17g/mol}=14.12mol$

Now we have to calculate the moles of $N_2$

The balanced chemical equation is:

$N_2+3H_2\rightarrow 2NH_3$

From the reaction, we conclude that

As, 2 mole of $NH_3$ produces from 1 mole of $N_2$

So, 14.12 mole of $NH_3$ produces form $\frac{14.12}{2}=7.06$ mole of $N_2$

Now we have to calculate the mass of $N_2$

$\text{ Mass of }N_2=\text{ Moles of }N_2\times \text{ Molar mass of }N_2$

Molar mass of $N_2$ = 28 g/mole

$\text{ Mass of }N_2=(7.06moles)\times (28g/mole)=197.68g\approx 198g$

Therefore, the mass of $N_2$ required are, 198 grams.