Answer to that question 

Asked by bpradeksha | 12th Feb, 2019, 07:54: PM

Expert Answer:

Given:
 
Volume of CO = 450 ml
 
Volume of O2 = 200 ml
 
The reaction is,
 
    space 2 CO subscript open parentheses straight g close parentheses end subscript space plus space space straight O subscript 2 subscript open parentheses straight g close parentheses end subscript end subscript space space space space space rightwards arrow space space space 2 CO subscript 2 subscript open parentheses straight g close parentheses end subscript end subscript
2 space vol space space space space space space space space space space 1 vol space space space space space space space space space space space space 2 space vol

The space amount space of space unused space CO comma

CO space colon space straight O subscript 2
space space space 2 space colon space 1
space space space space straight x space colon space 200
space space space space straight x space equals space 400 space ml

therefore space The space amount space of space unused space CO space equals space 450 space minus space 400 space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 50 space ml

The space amount space of space CO subscript 2 space formed comma
straight O subscript 2 space colon space space space CO subscript 2
space space 1 space colon space 2
200 colon space straight y

straight y space equals space 400 space ml

therefore space The space amount space of space space CO subscript 2 space formed space equals space 400 space ml
 
The composition of the resulting mixture is,
 
amount of unused CO = 50 ml
 
amount of CO2 formed = 400 ml. 

Answered by Varsha | 13th Feb, 2019, 11:40: AM