Question

56g of nitrogen and 8g of hydrogen gas are heated in a closed vessel. At equilibrium ,34g ammoniaare present. The equilibrium no. Of moles of n2 h2 and ammonia

Answer 1

3 H2 + N2 <======> 2 N H3
(4-3 x) ..(2- x )............ 2 x

At the start:
Moles of N2: 56/28 = 2.
moles of H2: 8/2= 4.

At equilibrium, 2 x = 34/17 =2 moles as given. So 3.4 moles of Ammonia are present.

So x = 1.0.
Moles of H2 and N2 present: 4-3x = 1 moles and 2 -x = 1 moles respectively.

Answer 2

Answer : The equilibrium number of moles of $N_2,H_2\text{ and }NH_3$ are, 1,1 and 2 respectively.

Solution : Given,

Mass of $N_2$ = 56 g

Mass of $H_2$ = 8 g

At eqm. the mass of $NH_3$ = 34 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the moles of $N_2,H_2\text{ and }NH_3$.

$\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{56g}{28g/mole}=2moles$

$\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{8g}{2g/mole}=4moles$

$\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}=\frac{34g}{17g/mole}=2moles$

Now we have to calculate the equilibrium number of moles of $N_2,H_2\text{ and }NH_3$.

The balanced chemical reaction will be,

                           $N_2+3H_2\rightarrow 2NH_3$

initially moles     2       4             0

At eqm.           $(2-\frac{x}{2})$  $(4-\frac{3x}{2})$     x

And as the number of moles of ammonia at equilibrium is, 2. That means, x = 2

Now put the value of 'x', we get the number of moles of hydrogen and nitrogen at equilibrium.

Number of moles of $N_2$ = $(2-\frac{x}{2})=(2-\frac{2}{2})=1$

Number of moles of $H_2$ = $(2-\frac{3x}{2})=(2-\frac{3\times 2}{2})=1$

Therefore, the equilibrium number of moles of $N_2,H_2\text{ and }NH_3$ are, 1,1 and 2 respectively.