Question

A 20.0 cm³mixture of CO, CH₄ and He gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be 13.0 cm³. A further contraction of 14.0 cm³ occurs when the residual gas is treated with KOH solution. Find out the composition of CO in initial gaseous mixture in terms of volume percentage.

Answer 1

Given :-

• A 20.0 cm³ mixture of CO, CH₄ and He gases is exploded by an electric discharge at room temperature with excess of oxygen.
• The volume contraction is found to be 13.0 cm³.
• A further contraction of 14.0 cm³ occurs when the residual gas is treated with KOH solution.

To Find :-

The composition of CO in initial gaseous mixture in terms of volume percentage.

Solution :-

$\sf{CO+\dfrac{1}{2}O_2 \longrightarrow CO_2}$

$\sf{x}$          $\sf{z}$

$\sf{O\:\:\:\:\:\:\:z-\dfrac{x}{2}-2y\:\:\:\:\:\:\:x}$

$\sf{CH_4+2O_2\longrightarrowCO_2+2H_2O _{(l)}}$

$\sf{y}$          $\sf{z}$

$\sf{O\:\:\:\:\:\:\:z-2y-\dfrac{x}{2}\:\:\:\:\:\:\:x}$

  $\sf{He \longrightarrow No\ reaction}$

$\sf{20-x-y}$

Volume of CO₂ = x + y = 14 cm³          . . . (i)

Volume contradiction = $\sf{v_i-v_f}$ = 13 cm³

$\sf{=20+z-(20-x-y+x+z-\dfrac{x}{2}-2y)}$

$\sf{\implies \dfrac{x}{2}+2y=13}$

$\sf{\therefore x+4y=26}$                                         . . . (ii)

From equations (i) and (ii),

x = 10, y = 4

_____________________

Now,

$\sf{Volume\ percentage\ of\ CO = \dfrac{x}{20} \times 100}$

$\sf{\implies Volume\ percentage\ of\ CO = \dfrac{10}{20} \times 100}$

$\bf{\implies Volume\ percentage\ of\ CO =50 \%}$

Therefore, the answer is 50·00.

Answer 2

Answer:

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