20 ml mixture of CH4 and CO2 when reacted with just sufficient oxygen,total volume after the complete reaction becomes 40 ml at 1 atm and 500 K (for all volume measurments)% of CH4 by volume in the initial mixture is
(A) 50%    (B) 100%    (C) 40%     (D) 70%

Dear Student,

The computation is as expressed below,

for reactant,n=PVRT=1×2010000.082×500=0.000487805 molafter reaction,n=1×4010000.082×500=0.00097561 mol% of methane = 0.0004878050.00097561×100%=50%Hence answer is Option (A) 50%

Regards.

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