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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