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ICSE Class 10 Answered

Plz solve the numerical
question image
Asked by lovemaan5500 | 14 Jan, 2018, 09:44: AM
answered-by-expert Expert Answer
The space combustion space reaction space of space water space gas space is space as

CO space space plus space straight H subscript 2 space plus space straight O subscript 2 space space rightwards arrow space space CO subscript 2 space space plus space space straight H subscript 2 straight O space

The space limiting space reagent space upon space which space the space concentration space of space carbon space dioxide space depends

Volume space of space water space gas space equals space 100 space cc

Volume space of space CO space equals space 50 space cc
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 0.5 space litre

Volume space of space space space straight H subscript 2 space space equals space 50 space cc
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 0.5 space litre

Volume space of space straight O subscript 2 space equals space 100 space cc space equals space 0.1 space litre

From space balanced space equation comma

CO space space plus space straight H subscript 2 space plus space straight O subscript 2 space space rightwards arrow space space CO subscript 2 space space plus space space straight H subscript 2 straight O space

22.4 space straight L space of space space straight O subscript 2 space reacts space with space 22.4 space straight L space of space CO

therefore space 0.1 space straight L space of space straight O subscript 2 space will space react space with space space fraction numerator 0.1 cross times 22.4 over denominator 22.4 end fraction

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 space space space space space equals space 0.1 space straight L space of space CO space

As space the space required space amount space of space CO space is space more space than space the space given space amount space hence comma space CO space is space the space limiting space reagent.

22.4 space straight L space of space CO space forms space 22.4 space straight L space CO subscript 2

therefore space 0.05 space straight L space of space CO space will space form space space space fraction numerator 0.05 cross times 22.4 over denominator 22.4 end fraction
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 space space space equals space 0.05 space straight L space of space CO subscript 2

therefore space The space amount space of space CO subscript 2 space formed space equals space 50 space cc

Now comma space the space amount space of space straight O subscript 2 space left space unreacted space equals space 100 minus 50 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 space space space space space space space space space space space space space space space space space space space equals space 50 space cc

As comma space 22.4 space straight L space of space space straight H subscript 2 space reacts space with space 22.4 space straight L space of space straight O subscript 2
therefore space 0.05 space straight L space of space straight H subscript 2 space will space react space with space space fraction numerator 0.05 cross times 22.4 over denominator 22.4 end fraction

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 space space space space space space space space space space space equals space 0.05 space straight L space of space straight O subscript 2

therefore space 50 space cc space of space space straight O subscript 2 space will space react space with space 50 space cc space of space space straight H subscript 2 space to space form space water.
Therefore space the space mixture space will space contain space 50 space cc space CO subscript 2 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
Answered by Varsha | 15 Jan, 2018, 12:49: PM
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