Ethene undergoes incomplete combustion to form carbon dioxide, carbon monoxide and water vapour. If the ratio of carbon dioxide to carbon monoxide formed is 9:1 when x cm^3 of ethene is burnt, what is the volume of oxygen gas consumed in the reaction?

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Ethene undergoes incomplete combustion to form carbon dioxide, carbon monoxide and water vapour. If the ratio of carbon dioxide to carbon monoxide formed is 9:1 when x cm^3 of ethene is burnt, what is the volume of oxygen gas consumed in the reaction?

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Answer 1 · 2017-05-29T07:59:00

Ethene undergoes incomplete combustion to form carbon dioxide, carbon monoxide and water vapour. The balanced equation of this incomplete combustion reaction is as follows.

$C_{2} H_{4} (g) + \frac{5}{2} O_{2} (g) \to C O_{2} (g) + C O (g) + 2 H_{2} O (g)$ $C_2H_4(g)+5/2O_2(g)->CO_2(g)+CO(g)+2H_2O(g)$

But as per question the ratio of volumes of $C O_{2} (g) : C O (g)$ $CO_2(g):CO(g)$ produced is $9 : 1$ $9:1$ . So this also represents the mole ratio of
$C O_{2} (g) : C O (g)$ $CO_2(g):CO(g)$ produced.

Hence adjusting this ratio balanced equation may be written as follows
$10 C_{2} H_{4} (g) + \frac{39}{2} O_{2} (g) \to 9 C O_{2} (g) + C O (g) + 20 H_{2} O (g)$ $10C_2H_4(g)+39/2O_2(g)->9CO_2(g)+CO(g)+20H_2O(g)$

Here we see $10 c m^{3}$ $10cm^3$ $C_{2} H_{4} (g)$ $C_2H_4(g)$ will require $\frac{39}{2} = 19.5 c m^{3} O_{2} (g)$ $39/2=19.5cm^3"" ""O_2(g)$

So to burn $x c m^{3}$ $xcm^3$ ethene the required volume of oxygen will be $\frac{19.5}{10} \times x c m^{3} = 1.95 x c m^{3}$ $19.5/10xx xcm^3=1.95xcm^3$

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Ethene undergoes incomplete combustion to form carbon dioxide, carbon monoxide and water vapour. If the ratio of carbon dioxide to carbon monoxide formed is 9:1 when x cm^3 of ethene is burnt, what is the volume of oxygen gas consumed in the reaction?

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