When CO$$_{2(g)}$$ is passed over red hot coke it partially gets reduced to CO$$(g)$$. Upon passing $$0.5$$ L of CO$$_2(g)$$ over red hot coke, the total volume of the gases increased to $$700$$ mL. The composition of the gaseous mixture at STP is

The reduction of carbon dioxide by red-hot coke follows the reaction

$$CO_{2(g)} + C_{(s)} \rightarrow 2\,CO_{(g)}$$

Because the reaction takes place at the same temperature and pressure (STP), gas volumes are directly proportional to the number of moles.

Let the volume of $$CO_2$$ that actually reacts be $$x$$ mL. Initial volume of $$CO_2$$ = $$500$$ mL.

Using the stoichiometry of the equation: $$1$$ volume of $$CO_2$$ $$\rightarrow$$ $$2$$ volumes of $$CO$$. Therefore, volume of $$CO$$ produced = $$2x$$ mL.

Volumes present after the reaction:   • $$CO_2$$ left = $$500 - x$$ mL   • $$CO$$ formed = $$2x$$ mL

The total final volume is given to be $$700$$ mL.

Hence, $$(500 - x) + 2x = 700$$ 

$$500 + x = 700$$ 

$$x = 200\ \text{mL}$$

Substituting $$x = 200$$ mL:

Remaining $$CO_2 = 500 - 200 = 300$$ mL Produced $$CO = 2 \times 200 = 400$$ mL

Thus, the composition of the gaseous mixture at STP is

$$CO_2 = 300\ \text{mL}; \qquad CO = 400\ \text{mL}$$

Option A which is: $$CO_2 = 300$$ mL; $$CO = 400$$ mL