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Question 43

5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is

We recall that any straight-chain or branched alkane has the general molecular formula $$C_nH_{2n+2}$$. For the complete combustion of such an alkane in oxygen, the balanced chemical equation (with water taken in the gaseous form so that every species can be treated as a gas at the given temperature and pressure) is first written as

$$C_nH_{2n+2}+O_2 \rightarrow n\,CO_2+\frac{n+1}{2}\,H_2O$$

Now we must balance the oxygen atoms. On the right side we have $$2n$$ oxygen atoms in $$n\,CO_2$$ and $$\frac{n+1}{2}\times 1$$ oxygen atoms in $$\frac{n+1}{2}\,H_2O$$. Adding them, the total number of oxygen atoms required is

$$2n+\frac{n+1}{2}=2n+\frac{n}{2}+\frac{1}{2}=\frac{5n+1}{2}.$$

Because one molecule of $$O_2$$ contributes two oxygen atoms, the number of $$O_2$$ molecules needed is half of this value. Therefore, the correctly balanced equation becomes

$$C_nH_{2n+2}+\frac{3n+1}{2}\,O_2\;\rightarrow\;n\,CO_2+\frac{n+1}{2}\,H_2O.$$

The coefficient in front of $$O_2$$ is clearly $$\dfrac{3n+1}{2}$$.

At constant temperature and pressure, equal volumes of gases contain equal numbers of moles (Gay-Lussac’s law of combining volumes). Hence the ratio of the coefficients in the balanced equation gives us directly the ratio of the volumes consumed. Thus, the volume ratio

$$\frac{\text{Volume of }O_2}{\text{Volume of alkane}}=\frac{3n+1}{2}.$$

According to the data, $$5\,\text{L}$$ of the alkane need $$25\,\text{L}$$ of oxygen. So the experimental ratio is

$$\frac{25}{5}=5.$$

Equating the theoretical and experimental ratios, we get

$$\frac{3n+1}{2}=5.$$

Multiplying both sides by $$2$$ to clear the denominator,

$$3n+1=10.$$

Subtracting $$1$$ from both sides,

$$3n=9.$$

Finally, dividing by $$3$$,

$$n=3.$$

Thus the alkane contains three carbon atoms, and its molecular formula is obtained by substituting $$n=3$$ into $$C_nH_{2n+2}$$:

$$C_3H_8.$$

$$C_3H_8$$ is propane. Checking the options, propane corresponds to Option D.

Hence, the correct answer is Option D.

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