The correct option is C 440 C_4H_10+132O_2→ KO_2+5H_2O CO+12O_2→ CO_2↑ H_2O ∴ total volume of CO_2 produced =4× volume of C_4H_10 ...

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Measurement of Volume

The

Question

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.

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 440
$C_{4} H_{10} + \frac{13}{2} O_{2} \to K O_{2} + 5 H_{2} O$
$C O + \frac{1}{2} O_{2} \to C O_{2} ↑ H_{2} O$
$∴$ total volume of $C O_{2}$ produced
$= 4 \times v o l u m e o f C_{4} H_{10}$
$= 1 \times (v o l u m e o f C O)$
$= 4 \times 80 \times 120$
$= 440 m l$

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Similar questions

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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 O4 produced.

The correct option is C 440C4H10+132O2→KO2+5H2OCO+12O2→CO2↑H2O∴ total volume of CO2 produced=4×volumeofC4H10=1×(volumeofCO)=4×80×120=440ml

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.

The correct option is C 440
$C_{4} H_{10} + \frac{13}{2} O_{2} \to K O_{2} + 5 H_{2} O$
$C O + \frac{1}{2} O_{2} \to C O_{2} ↑ H_{2} O$
$∴$ total volume of $C O_{2}$ produced
$= 4 \times v o l u m e o f C_{4} H_{10}$
$= 1 \times (v o l u m e o f C O)$
$= 4 \times 80 \times 120$
$= 440 m l$