mixture of co2 n co is passed over red hot graphite when 1 mole of mixture changes to 33.6l(stp)the mole fraction of co2 in the mixture is?

Assume that the mixture contains y mol of CO₂ and (1 - y) mol of CO.

When a mixture of CO₂ and CO is passed over red hot graphite. CO does not react, while CO₂ is reduced to CO.
CO₂(g) + C(s) → 2CO(g)
y mol of CO₂ is converted to 2y mol of CO.

1 mole of gas at STP occupies a volume of 22.4 L.
No. of moles of the gas mixture after reaction:
2y + (1 - y) = 33.6/22.4
2y + 1 - y = 1.5
y + 1 = 1.5
y = 0.5

Mole fraction of CO₂ in the mixture = 0.5/1 = 1/2