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Question

A mixture of CH4 & O2 is used as an optimal fuel if O2 is present in thrice the amount required theoretically for combustion of CH4. Calculate number of effusions steps required to convert a mixture containing 1 part of CH4 in 193 parts mixture (parts by volume). If calorific value (heat evolved when 1 mole is burnt) of CH4 is 100 cal/mole & if after each effusion 80% of CH4 is collected. Find out what initial mole of each gas in initial mixture required for producing 1000 cal of energy after processing.
[Given (0.9)5=0.6]

A
9Steps,14.32molCH4,2886.7molO2
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B
10Steps,13.89molCH4,2666.65molO2
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C
10Steps,27.78molCH4,5333.3molO2
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D
9Steps,28.64molCH4,5773.4molO2
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Solution

The correct option is C 10Steps,27.78molCH4,5333.3molO2
CH4+2O2CO2+H2O
3 times O2 theoretical is required for optimal fuel
For optimal fuel nCH4nO2=12×3=16
We know,
rCH4rO2=nCH4nO2×MO2MCH4
16=1192×(2)n; 1926=(2)n
32=(2)n
n= 10 steps
Let initial moles be nA,nB
Afer 1 effusion
nAnb=nAinBi×2
And 90% 0f nA is remove
nAnO=nAinBi×2×0.9
After 10 steps
We know
1 mole CH4 produces 100 - fours
to produces
13=nAinBi×(2x)
16=nAinBi×32x
Also o produce 1000 cal,10 moles of CH4 is begin
10=nCH4×0.910
nCH4=27.78 mole
MOles of O2=2778×3×36×32=5333.33 moles

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