Question

One mole of $O_2$ (g) was passed over hot coke. At the end of the reaction, 40 % of $O_2$ (g) was unreacted. What is the volume (in L) of reaction mixture at STP (273.15 K and 1 bar)? (Assume only CO (g) is formed in the reaction)

• A. 22.7
• B. 72.64
• C. 36.32
• D. 45.4

Hint:

Calculate remaining reactants and product moles, then use molar volume at STP.

Answer 1

The Correct Option is C

Step 1: Write the balanced reaction.
Hot coke (carbon) with oxygen, making only carbon monoxide, follows \[ O_2(g) + 2C(s) \rightarrow 2CO(g). \] So one mole of $O_2$ makes two moles of $CO$ when fully reacted.

Step 2: Find how much oxygen reacted.
We start with 1 mole of $O_2$. Since 40 percent is left unreacted, the leftover is $0.4$ mole and the reacted part is $0.6$ mole.

Step 3: Find the CO formed.
From the equation, each mole of reacted $O_2$ gives two moles of $CO$. So $CO = 2 \times 0.6 = 1.2$ moles.

Step 4: Total up the gas moles.
The final gas mixture is the leftover oxygen plus the new carbon monoxide: \[ n_{total} = 0.4 + 1.2 = 1.6 \text{ moles}. \]

Step 5: Use the molar volume at this STP.
At 273.15 K and 1 bar, one mole of gas takes about $22.7$ L. So \[ V = 1.6 \times 22.7 = 36.32 \text{ L}. \]

Step 6: State the answer.
The volume of the reaction mixture is about $36.32$ L.
\[ \boxed{36.32} \]