Question

When \[C{{O}_{2}}\](g)is passed over red-hot coke it partially gets reduced to \[CO\](g). Upon passing 0.5 litre of \[C{{O}_{2}}\](g) over red hot coke, the total volume of the gases increased to 700 mL . The composition of the gaseous mixture at STP is:
A. \[C{{O}_{2}}\] = 200 mL; \[CO\] = 500 mL
B. \[C{{O}_{2}}\] = 350 mL; \[CO\] = 350 mL
C. \[C{{O}_{2}}\] = 0.0 mL; \[CO\] = 700 mL
D. \[C{{O}_{2}}\] = 300 mL; \[CO\] = 400 mL

Answer 1

Hint: The equation for the red hot coke is \[C{{O}_{2}}+C\to 2CO\]. The constant value to convert gas at STP is 22.4. By dividing the gas volume by this constant find the volume of gas at STP.

Complete step by step answer:
Let us assume that ‘x’ liter of \[C{{O}_{2}}\]is consumed in the reaction.
We know,

22.4L of \[C{{O}_{2}}\] = 1mole of \[C{{O}_{2}}\] so

X L of \[C{{O}_{2}}\] = \[\dfrac{x}{22.4}\]moles of \[C{{O}_{2}}\] is consumed.

The moles of \[C{{O}_{2}}\] left after the reaction will be moles of\[(\dfrac{0.5-x}{22.4})\]

From the equation we get to know that 1 mole of \[C{{O}_{2}}\] forms 2 moles of \[CO\] so,

\[\dfrac{x}{22.4}\]moles of \[C{{O}_{2}}\]will form= \[2(\dfrac{x}{22.4})\] moles of \[CO\]

We have to find the composition of gaseous mixture at STP. So, the composition at STP will be the amount of \[C{{O}_{2}}\] left and the amount of \[CO\] formed.
The total volume increased is =700mL (0.7L)
The total volume increased in moles will be

22.4L of volume = 1 mole of the substance so,

0.7L of volume = \[\dfrac{0.7}{22.4}\]moles of the substance

The composition of mixture will be = left amount of \[C{{O}_{2}}\] moles + \[CO\]formed moles

\[\dfrac{0.7}{22.4}\] moles = \[(\dfrac{0.5-x}{22.4})\] + \[2(\dfrac{x}{22.4})\]

22.4 in the denominator will get cancel and we will get

0.7= 0.5-x +2x

x = 0.2L (200mL)

So the volume of \[C{{O}_{2}}\] consumed is 200mL and \[CO\] formed is 400mL (2x of \[C{{O}_{2}}\] consumed)

Now the \[C{{O}_{2}}\]left = given volume-consumed volume

= 500mL-200mL

=300mL

The left \[C{{O}_{2}}\]is 300mL and formed \[CO\] is 400mL.

So, the correct answer is Option D.

Note: The students generally do not convert the given volumes in moles and get confused while solving them. They also find the left \[C{{O}_{2}}\]by taking a variable and making the question. So students must read the question thoroughly and understand it.