The volume of the gas mixture obtained by passing 400 mL `CO_(2)` over red hot coke was measured to be 600 mL. find the compo-sition of the gas mixture.

Explanation

Reaction: `C(s)+CO_(2)(g) to 2CO(g)`. Let as consider, 400 mL `CO_(2)` gas on reduction produce x mL CO gas.
According to the above equation, x mL `CO_(2)-=2x" mlCO"`
Volume of residual `CO_(2)` in the mixture=(400-x) mL
total volume of the mixture
`=(400-x+2x)mL=(400+x)mL`
As given in the question, 400+x=600
`therefore x=200`
`therefore` 200 mL of `CO_(2) and 2xx200 mL=400mL` of CO are present.