Science, asked by sy083408, 10 months ago

40 ml of a mixture of ch4 and co2 undergoes complete combustion with 80 ml of o2. X cal of reduction in volume were observed when the gaseous mixture was passed through alkaline pyrogallol 20 ml of reduction was absorbed the value of X and volume of co2 in the original mixture will be

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Answered by Anonymous

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Let volume of CO, CH4 and Ne be x, y and z respectively.

CO + 1/2O2  CO2 .....(1)

here it is clear that x volume of CO reacts with x/2 volume of O2 and produces sx volume of CO2

and CH4 + 2O2  CO2 + 2H2O (l).....(2)

here it is clear that y volume of CH4 reacts with 2y volume of O2 and produces y volume of CO2.

now remaining volume of O2 = 10 - x/2 - 2y

volume after reaction,

volume of CO2 [eq(1)] + volume of CO2 [eq(2)] + volume of oxygen remaining + volume of Ne = 36.5 ml

or, x + y + 10 - x/2 - 2y + z = 36.5 ml

or, x/2 - y + z = 26.5 ml ......(i)

given, volume of Co + volume of CH4 + volume of Ne = 40ml

or, x + y + z = 40ml.....(ii)

we know, KOH reduces CO2

so, volume of CO2 [eq(1) ] + volume of CO2 [eq(2)] = 9ml

or, x + y = 9ml.....(iii)

solving equations (i), (ii) and (iii)

z = 31 , x = 3 and y = 6

so, volume of CH4 = 6ml

% of CH4 in the original mixture = 6/40 × 100 = 15%. $<body bgcolor="yellow">$

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Answered by Btwitsaditi12

hey mate, your answer :

Let volume of CO, CH4 and Ne be x, y and z respectively.

CO + 1/2O2  CO2 .....(1)

here it is clear that x volume of CO reacts with x/2 volume of O2 and produces sx volume of CO2

and CH4 + 2O2  CO2 + 2H2O (l).....(2)

here it is clear that y volume of CH4 reacts with 2y volume of O2 and produces y volume of CO2.

now remaining volume of O2 = 10 - x/2 - 2y

volume after reaction,

volume of CO2 [eq(1)] + volume of CO2 [eq(2)] + volume of oxygen remaining + volume of Ne = 36.5 ml

or, x + y + 10 - x/2 - 2y + z = 36.5 ml

or, x/2 - y + z = 26.5 ml ......(i)

given, volume of Co + volume of CH4 + volume of Ne = 40ml

or, x + y + z = 40ml.....(ii)

we know, KOH reduces CO2

so, volume of CO2 [eq(1) ] + volume of CO2 [eq(2)] = 9ml

or, x + y = 9ml. (iii)

solving equations (i) (ii) and (iii)

z = 31 , x = 3 and y = 6

so, volume of CH4 = 6ml

% of CH4 in the original mixture = 6/40 × 100 = 15%

hope it helps u and Mark as the brainliest..

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40 ml of a mixture of ch4 and co2 undergoes complete combustion with 80 ml of o2. X cal of reduction in volume were observed when the gaseous mixture was passed through alkaline pyrogallol 20 ml of reduction was absorbed the value of X and volume of co2 in the original mixture will be​

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40 ml of a mixture of ch4 and co2 undergoes complete combustion with 80 ml of o2. X cal of reduction in volume were observed when the gaseous mixture was passed through alkaline pyrogallol 20 ml of reduction was absorbed the value of X and volume of co2 in the original mixture will be