The correct option is A 173ml C_3H_8+5O_2→ 3CO_2+4H_2O CH_4+2O_2→ CO_2+2H_2O CO+12O_2→ CO_2 Volume of CO_2 produced from C_3H_8=3×36.5= ...

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Question

The % by volume of $C_{3} H_{8}$ in a mixture of $C_{3} H_{8}$ , $C H_{4}$ and $C O$ is 36.5 . Calculate the volume of $C O_{2}$ produced when 100 ml of the mixture is burnt in excess of $O_{2}$ .

173ml

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237ml

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475ml

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650ml

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Solution

The correct option is A 173ml
$C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
$C O + \frac{1}{2} O_{2} \to C O_{2}$
Volume of $C O_{2}$ produced from $C_{3} H_{8} = 3 \times 36.5 = 109.5 m l$
Volume of $C O_{2}$ produced from $C H_{4} a n d C O = 1 \times 63.5 = 63.5 m l$
Total volume of $C O_{2} = 63.5 + 109.5 = 173 m l$

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Similar questions

The percentage by volume of $C_{3} H_{8}$ in a mixture of $C_{3} H_{8}$ , $C H_{4}$ and CO is 36.5. Calculate the volume of $C O_{2}$ produced when 100 mL of mixture is burnt in excess of $O_{2}$ .

Q. The percentage by volume of

C_{3} H_{8}

in a mixture of

C_{3} H_{8}, C H_{4}

and

C O

36.5

. The volume of

C O_{2}

produced (in mL), when

100

mL of the mixture is burnt in excess of

O_{2}

, is :

Q. The percentage volume of

C_{3} H_{8}

in a mixture of

C_{3} H_{8}, C H_{4}

and

C O

36.5

. The volume of

C O_{2}

produced when

100

mL of mixture is burnt in excess of

O_{2}

is:

Q. The

%

by volume of

C_{4} H_{10}

in a gaseous mixture of

C_{4} H_{10}

C H_{4}

and CO is

40

. When

200

ml of the mixture is burnt in excess of

O_{2}

. Find volume (in ml) of

C O_{2}

produced.

Q. The % by volume of

C_{4} H_{10}

in a gaseous mixture of

C_{4} H_{10}

and CO is 40. When 200 ml of the mixture is burnt in excess of

O_{2}

. Find volume (in ml) of

C O_{2}

produced.

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The % by volume of C3H8 in a mixture of C3H8, CH4 and CO is 36.5 . Calculate the volume of CO2 produced when 100 ml of the mixture is burnt in excess of O2 .

The correct option is A 173mlC3H8+5O2→3CO2+4H2OCH4+2O2→CO2+2H2OCO+12O2→CO2Volume of CO2 produced from C3H8=3×36.5=109.5 mlVolume of CO2 produced from CH4 and CO=1×63.5=63.5 mlTotal volume of CO2=63.5+109.5=173 ml

The % by volume of $C_{3} H_{8}$ in a mixture of $C_{3} H_{8}$ , $C H_{4}$ and $C O$ is 36.5 . Calculate the volume of $C O_{2}$ produced when 100 ml of the mixture is burnt in excess of $O_{2}$ .