wht would be percentage composition by volume of a mixture of CO and CH4 whose 10.5 ml requires 9 ml oxygen for combution?1)CO=80% CH4=20% 2)CO =90% CH4 =10%3) CO=76.2% CH4=23.8% 4) CO=66% CH4=34%

Dear Soumya,

Moles = pV / RT = 1.00 atm x 1.00 L / 0.0821 x 273.15 = 0.0446

let x = moles CH4 and let y = moles C2H6

x + y = 0.0446
x ( 890.3) + y ( 1559.7) = 46.9

x = 0.0446 -y
39.7 - 890.3 y + 1559.7 y = 46.9

669.4 y = 7.2

y = 0.0108 moles C2H6
x = 0.0446 - 0.0108 = 0.0338

V(C2H6) = 0.0338 x 0.0821 x 273.15 / 1.00 =0.758 L
V ( CH4) = 0.0108 x 0.0821 x 273.15 / 1.00 =0.242 L

% by volume = 0.242 x 100 / 1.00 = 24.2 ( CH4)
and 75.8 % C2H6

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