NEET · Chemistry · STD 11 - 1. Some basic concept of chemistry
When \(1 dm ^3\) of \(CO _2\) gas is passed over hot coke, the volume of gaseous mixture after complete reaction at STP becomes \(1.4 dm ^3\). The composition of the gaseous mixture at STP is :
- A \(0.8 dm ^3\) of \(CO , 0.8 dm ^3\) of \(CO _2\)
- B \(0.8 dm ^3\) of \(CO , 0.6 dm ^3\) of \(CO _2\)
- C \(0.6 dm ^3\) of \(CO , 0.8 dm ^3\) of \(CO _2\)
- D \(0.6 dm ^3\) of \(CO , 0.4 dm ^3\) of \(CO _2\)
Answer & Solution
Correct Answer
(B) \(0.8 dm ^3\) of \(CO , 0.6 dm ^3\) of \(CO _2\)
Step-by-step Solution
Detailed explanation
(B) The reaction of carbon dioxide with hot coke is given by:
\(CO_2(g)+C(s) \rightarrow 2 CO(g)\)
Initial volume of \(CO _2=1 dm ^3\)
Let \(x dm ^3\) of \(CO _2\) react with hot coke.
Volume of \(CO _2\) remaining \(=1-x dm ^3\)
Volume of CO formed \(=2 x dm ^3\)
Total volume of the gaseous mixture \(=(1-x)+2 x=1+x dm ^3\)
Given that the total volume after the reaction is \((A)4 dm ^3\) :
\(1+x=(A)4\)
\(x=0.4 dm^3\)
Volume of \(CO _2\) remaining \(=1-0.4=0.6 dm ^3\)
Volume of CO formed \(=2 \times 0.4=0.8 dm ^3\)
The composition of the gaseous mixture is \(0.8 dm ^3\) of CO and \(0.6 dm ^3\) of \(CO _2\).
\(CO_2(g)+C(s) \rightarrow 2 CO(g)\)
Initial volume of \(CO _2=1 dm ^3\)
Let \(x dm ^3\) of \(CO _2\) react with hot coke.
Volume of \(CO _2\) remaining \(=1-x dm ^3\)
Volume of CO formed \(=2 x dm ^3\)
Total volume of the gaseous mixture \(=(1-x)+2 x=1+x dm ^3\)
Given that the total volume after the reaction is \((A)4 dm ^3\) :
\(1+x=(A)4\)
\(x=0.4 dm^3\)
Volume of \(CO _2\) remaining \(=1-0.4=0.6 dm ^3\)
Volume of CO formed \(=2 \times 0.4=0.8 dm ^3\)
The composition of the gaseous mixture is \(0.8 dm ^3\) of CO and \(0.6 dm ^3\) of \(CO _2\).
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