The correct option is D 440 C_4H_10+132O_2→ KO_2+5H_2O CO+12O_2→ Cn_2↑ H_2O ∴ total volume of CO_2 produced =4× (volume of C_4H_ ...

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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 of $O_{2}$ . 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 D 440
$C_{4} H_{10} + \frac{13}{2} O_{2} \to K O_{2} + 5 H_{2} O$
$C O + \frac{1}{2} O_{2} \to C n_{2} ↑ H_{2} O$
$∴$ total volume of $C O_{2}$ produced
$= 4 \times$ (volume of $C_{4} H_{10})$
$\Rightarrow 1 \times$ (volume of $C O$ )
$= 4 \times 80 + 120$
$= 440 m l$

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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 of O2. Find volume (in ml) of CO2 produced.

The correct option is D 440C4H10+132O2→KO2+5H2OCO+12O2→Cn2↑H2O∴ total volume of CO2 produced=4× (volume of C4H10)⇒1× (volume of CO)=4×80+120=440ml

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.

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