Question

$10mL C_4H_x(g)$ requires $55mL O_2(g)$ for its complete combustion. Find the value of $x$.

• A. 4
• B. 6
• C. 8
• D. 10

Answer 1

Answer: (B)

The value of $x$ is 6.

Answer 2

The complete combustion of a hydrocarbon $C_aH_b$ is represented by the balanced chemical equation:

$C_aH_b(g) + (a + \frac{b}{4}) O_2(g) \rightarrow a CO_2(g) + \frac{b}{2} H_2O$

In this problem, the hydrocarbon is $C_4H_x$. So, the balanced chemical equation for the combustion of $C_4H_x$ is:

$C_4H_x(g) + (4 + \frac{x}{4}) O_2(g) \rightarrow 4 CO_2(g) + \frac{x}{2} H_2O$

According to Gay-Lussac's Law of Gaseous Volumes, when gases react at the same temperature and pressure, the ratio of their volumes is equal to the ratio of their stoichiometric coefficients in the balanced chemical equation.

From the balanced equation, 1 volume of $C_4H_x(g)$ reacts with $(4 + \frac{x}{4})$ volumes of $O_2(g)$.

We are given that $10mL$ of $C_4H_x(g)$ requires $55mL$ of $O_2(g)$ for complete combustion. Using the ratio of volumes:

$\frac{\text{Volume of } C_4H_x}{\text{Volume of } O_2} = \frac{\text{Stoichiometric coefficient of } C_4H_x}{\text{Stoichiometric coefficient of } O_2}$

$\frac{10 \text{ mL}}{55 \text{ mL}} = \frac{1}{(4 + \frac{x}{4})}$

Simplify the fraction on the left side:

$\frac{10}{55} = \frac{2}{11}$

So, the equation becomes:

$\frac{2}{11} = \frac{1}{4 + \frac{x}{4}}$

Cross-multiply:

$2 \times (4 + \frac{x}{4}) = 11 \times 1$

$8 + \frac{2x}{4} = 11$

$8 + \frac{x}{2} = 11$

Subtract 8 from both sides:

$\frac{x}{2} = 11 - 8$

$\frac{x}{2} = 3$

Multiply by 2:

$x = 3 \times 2$

$x = 6$

Thus, the value of $x$ is 6.