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Question

# The % by volume of C4H10 in a gaseous mixture of C4H10, CH4 and CO is 40. When 200 ml of the mixture is burnt in excess of O2. Find volume (in ml) of CO2 produced.

A
220
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B
340
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C
440
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D
560
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Solution

## The correct option is C 440% by volume of C4H10=40%Total volume of mixture=200mLVolume of C4H10=40100×200Volume of C4H10=80mLRemaining Volume=200−80=120mLVolume of CH4+CO=120mLLet Volume of CH4 be V mLthen volume of CO will be 120−VmLCH4+2O2⟶CO2+2H2OVmL VmLCO+12O2⟶CO2(120-V)mL (120-V)mLC4H10+132O2⟶4CO2+5H2O80ml 80x4=320mLTotal volume of CO2 produced=VmL+(120−V)mL+320mL=(V+120−V+320)mL=440mL

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