Question

When CO₂(g) is passed over red hot coke. It partially gets reduced to CO(g). Upon passing 0.5L of CO₂(g) over red hot coke. The total volume of the gases increased to 700mL. The composition of the gaseous mixture at STP is

A.CO₂ 200 mL;CO = 500 mL
B.CO₂ = 350 mL;CO = 350 mL
C.CO₂ = 300 mL;CO = 400 mL
D.CO₂ = 0.0 mL;CO = 700 mL

Answer 1

Upon passing 0.5L of CO₂ over red hot coke, the total gas volume increased to 700mL, indicating all CO₂ was converted to CO, resulting in a gaseous composition of CO₂ = 0.0 mL and CO = 700 mL at STP.

Step-by-step explanation:

When carbon dioxide (CO₂) is passed over red hot coke, a chemical reaction occurs in which the carbon dioxide is reduced to carbon monoxide (CO), according to the following balanced equation:

CO₂(g) + C(s) ⇒ 2 CO(g)

Starting with 0.5L of CO₂ and observing an increase in total volume up to 700mL after the reaction, we can apply the stoichiometry of the reaction to find the composition of the gaseous mixture. Since the reaction produces two moles of CO for every mole of CO₂ consumed and the volumes of gases under the same conditions are proportional to the number of moles, the volume of CO produced will be twice the volume of CO₂ that reacted. However, given that the total volume increased to 700mL, it implies that all of the CO₂ was converted to CO.

Thus, the composition of the gaseous mixture at STP is:

CO₂ = 0.0 mL; CO = 700 mL.