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Oxidation of Alkanes

The percentag...

Question

The percentage volume of $C_{3} H_{8}$ in a mixture of $C_{3} H_{8}, C H_{4}$ and $C O$ is $36.5$ . The volume of $C O_{2}$ produced when $100$ mL of mixture is burnt in excess of $O_{2}$ is:

273

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183

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173

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193

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Solution

The correct option is B $273$ mL
Let the volume of mixture of all gases be 100 unit.

The reaction of combustion as follows:
$C_{3} H_{8} + C H_{4} + C O + 15 / 2 O_{2} \to 5 C O_{2} + 6 H_{2} O$

It means 5 mol of $C O_{2}$ is produced when 1 mol $C_{3} H_{8}$ is burnt in a given mixture, along with 1 mol each of $C H_{4}$ and CO.

As 22400 mL of $C_{3} H_{8}$ contains 1 mol, then 36.5 mL contains = $\frac{1}{22400} \times 36.5$ moles

1 mol of $C_{3} H_{8}$ produces 3 mol of $C O_{2}$ (2 moles from $C H_{4}$ and $C O$ )

Then, $\frac{1}{22400} \times 36.5$ mol produces $\frac{36.5}{22400} \times 3$ moles of $C O_{2}$ .

Now $C H_{4}$ + CO combinely must have $(\frac{1}{22400} \times 100) - (\frac{1}{22400} \times 36.5)$ moles, in 100 mL mixture

Now as per the above molar equation for all gases of the mixture, mixture gives 5 mol of $C O_{2}$ .
so 1 mol of each $C H_{4} a n d C O$ must give 2 mol of $C O_{2}$ .

so from 100 mL mixture of $C H_{4} + C O$ will give

$= \frac{100}{22400} \times 2 - \frac{1}{22400} \times 36.5$ moles

Now total mol from the mixture, $C O_{2}$ moles produced

$= \frac{100}{22400} \times 2 - \frac{1}{22400} \times 36.5 + \frac{36.5}{22400} \times 3 = \frac{273}{22400} = 0.0121875 m o l e s$
As 1 mol $C O_{2}$ has 22400 mL volume.
So 0.0121875 mol have volume $= 0.0121875 \times 22400 = 273 m L$

Hence, option $A$ is correct.

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Similar questions

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The percentage by volume of $C_{3} H_{8}$ in a mixture of $C_{3} H_{8}$ , $C H_{4}$ and CO is 36.5. Calculate the volume of $C O_{2}$ produced when 100 mL of mixture is burnt in excess of $O_{2}$ .

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The percentage volume of C3H8 in a mixture of C3H8, CH4 and CO is 36.5. The volume of CO2 produced when 100 mL of mixture is burnt in excess of O2 is:

The percentage volume of $C_{3} H_{8}$ in a mixture of $C_{3} H_{8}, C H_{4}$ and $C O$ is $36.5$ . The volume of $C O_{2}$ produced when $100$ mL of mixture is burnt in excess of $O_{2}$ is: