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Volume of Gases and Number of Moles

A. mixture of...

Question

A. mixture of methane and ethylene in the volume ratio x:y has a total volume of 50m1. On complete combustion it gave 80m1 of ${C O}_{2}$ . If the ratio had been y:x, instead of x:y, what volume of ${C O}_{2}$ could have been obtained?,

50 ml

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70 ml

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90 ml

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75 ml

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Solution

The correct option is B 70 ml
$C H_{4} + 2 O_{2} ⟶ C O_{2} + 2 H_{2} O$ $X C H_{4} = X C O_{2}$
$Y C_{2} H_{4} = 2 Y C O_{2} \Rightarrow C_{2} H_{4} + 3 O_{2} ⟶ 2 C O_{2} + 2 H_{2} O$
$\Rightarrow X (C H_{4}) + Y (C_{2} H_{4}) ⟶ (x + 2 y) C O_{2}$

$x + y = 50 m l, x + 2 y = 80 m l$
$\Rightarrow x + y + y = 80 m l$
$\Rightarrow 50 + y = 80 m l$
$\Rightarrow y = 30 m l, x = 20 m l$ .

If ratio of methane & $C_{2} H_{4}$ are altered, methane $y = 30 m l$ & ethane $x = 20 m l$ .
Reactants= $x + y = 20 + 30 = 50 m l$
Product= $2 x + y = 2 (20) + 30 = 70 m l$ .
$∴ 70 m l$ $C O_{2}$ is produced .

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A. mixture of methane and ethylene in the volume ratio x:y has a total volume of 50m1. On complete combustion it gave 80m1 of CO2. If the ratio had been y:x, instead of x:y, what volume of CO2 could have been obtained?,

A. mixture of methane and ethylene in the volume ratio x:y has a total volume of 50m1. On complete combustion it gave 80m1 of ${C O}_{2}$ . If the ratio had been y:x, instead of x:y, what volume of ${C O}_{2}$ could have been obtained?,