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Question

$$40 \ mL$$ of a mixture of hydrogen, $$CH_4$$ and $$N_2$$ was exploded with $$10 \ mL$$ of oxygen. On cooling, the gases occupied $$36.5 \ mL$$. After treatment with $$KOH$$, the volume reduced by $$3 \ mL$$ and again on treatment with alkaline pyrogallol, the volume further decreased by $$1.5 \ mL$$. Determine the composition of the mixture.


Solution

$$ V(H_2 + CH_4 + N_2) = 40 mL$$

$$ V_{O_2} = 10 mL$$

$$ H_2 + \dfrac{1}{2} O_2  \rightarrow  \ H_2O(l)$$

$$ CH_4 + \left(1+\dfrac{4}{4}\right) O_2  \rightarrow \ CO_2 + 2H_2O(l)$$

$$ N_2 \rightarrow $$  No reaction 

Let volume of $$H_2 = x mL$$

Volume of $$CH_4 = y \ mL$$

volume of $$N_2 = \{40 - (x + y)\}$$

On cooling the gas remained 
$$= CO_2 (g) $$ produced $$ +  N_2(g)  +  O_2$$ (remained)

A/q, $$ y + \{40 - (x + y)\} + 10 - \left(\dfrac{x}{2} + 2y\right) = 36.5$$

(with x mL of $$H_2, \dfrac{x}{2}$$  mL of $$O_2$$ reacts & with y mL of $$CH_4 \ 2y \ mL$$ of $$O_2$$ reacts)

$$ 40 - x + 10 - \dfrac{x}{2} - 2y = 36.5$$

$$50 - \dfrac{3x}{2} - 2y = 36.5 $$

$$ 50 - 36.5 = \dfrac{3x}{2} + 2y$$

$$\therefore \dfrac{3x}{2} + 2y = 13.5$$

$$3x + 4y = 27$$ ---------------(1)

After passing over KOH, $$CO_2$$ will be absorbed
$$\therefore y = 3 \ ml$$ -------------(2)

After passing over pyrogallol, $$O_2$$ will be absorbed.

$$\therefore \ 10 - \left(\dfrac{x}{2} + 2y \right) = \dfrac{3}{2}$$

$$ 10 - \dfrac{3}{2} = \dfrac{x}{2} + 2y$$

$$ 17 = x + 4y$$ ------------------(3)

 from (2) & (3) $$x = 5$$

$$\therefore$$ Volume of $$H_2  = 5 mL$$

Volume of $$NH_4 = 3 mL$$

Volume of $$N_2 = 32 mL$$

$$\therefore \% \ of \ CH_4 = \dfrac{3}{40} \times  100 = 7.5\%$$

$$\therefore \% \ of \ H_2 = \dfrac{5}{40} \times  100 =12.5\%$$

$$\therefore \% \ of \ N_2 = \dfrac{32}{40} \times  100 = 80\%$$

Chemistry

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