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Asked by bpradeksha | 12 Feb, 2019, 19:54: PM
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](https://images.topperlearning.com/topper/tinymce/cache/d9b24bb1385d1bc1f556acf7b047715e.png)
The composition of the resulting mixture is,
amount of unused CO = 50 ml
amount of CO2 formed = 400 ml.
Answered by Varsha | 13 Feb, 2019, 11:40: AM
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