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Standard XII

Chemistry

Limiting Reagent or Reactant

The

Question

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 $O_{2} .$ Find volume(in ml) of $C O_{2}$ produced.

220

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340

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440

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560

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Solution

The correct option is C 560
As after reaction, each mole of CO will produce same no. of mole of $C O_{2}$ and each mole of $C_{4} H_{10}$ will form 4 times $C O_{2}$ so total volume of $C O_{2}$ formed is - $80 + 4 \times 120 = 560$ ml.

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Similar questions

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.

Q. The % by volume of

C_{4} H_{10}

in a gaseous mixture of

C_{4} H_{10}, C H_{4} a n d C O

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

O_{4}

produced.

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 % by volume of

C_{3} H_{8}

in a mixture of

C_{3} H_{8}

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and

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is 36.5 . Calculate the volume of

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produced when 100 ml of the mixture is burnt in excess of

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

The correct option is C 560As after reaction, each mole of CO will produce same no. of mole of CO2 and each mole of C4H10 will form 4 times CO2 so total volume of CO2 formed is - 80+4×120=560 ml.

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 $O_{2} .$ Find volume(in ml) of $C O_{2}$ produced.

The correct option is C 560
As after reaction, each mole of CO will produce same no. of mole of $C O_{2}$ and each mole of $C_{4} H_{10}$ will form 4 times $C O_{2}$ so total volume of $C O_{2}$ formed is - $80 + 4 \times 120 = 560$ ml.