Question

A gas mixture of 3.0 L of propane and butane on complete combustion at 25 degree Celsius produced 10 L of CO2. Find the composition of the gas mixture?

Answer 1

Answer on Question #44641 – Chemistry – Organic Chemistry

Question

A gas mixture of 3.0 L of propane and butane on complete combustion at 25 degree Celsius produced 10 L of CO₂. Find the composition of the gas mixture?

Solution

Chemical equations of propane and butane complete combustion are as follows

$$\mathrm{C_3H_8} + 5\mathrm{O_2} \rightarrow 3\mathrm{CO_2} + 4\mathrm{H_2O}$$ $$\mathrm{C_4H_{10}} + 6.5\mathrm{O_2} \rightarrow 4\mathrm{CO_2} + 5\mathrm{H_2O}$$

Let us assign the volume of propane in the initial mixture $V_p = x$, and the volume of butane $V_b = y$. It is given that $V_p + V_b = 3.0 \, \text{L}$, so

$$x + y = 3$$

As is clear from the chemical equation of propane combustion, volume of CO₂ produced at complete combustion of propane is 3 times greater than volume of propane V_{\mathrm{CO_2}^{(p)}} = 3 \, \text{V_p}. And from the chemical equation of butane combustion it is clear that V_{\mathrm{CO_2}^{(b)}} = 4 \, \text{V_b}.

It is given that $V_{\mathrm{CO_2}^{(p)}} + V_{\mathrm{CO_2}^{(b)}} = 10 \, \text{L}$, so

$$3x + 4y = 10$$

Thus, we have the set of two linear equations

$$\left\{ \begin{array}{l} x + y = 3 \\ 3x + 4y = 10 \end{array} \right.$$

The set of equations may be solved e.g. by substitution method:

$$\begin{array}{l} y = 3 - x \\ 3x + 4(3 - x) = 10 \\ 3x + 12 - 4x = 10 \\ 3x - 4x = 10 - 12 \\ x = 2 \\ y = 3 - 2 = 1 \\ \end{array}$$

Thus, 3 L of the propane-butane mixture contain 2 L of propane and 1 L of butane.

The gas mixture composition in % by vol.:

$$\% \text{propane} = \frac{2}{3} \cdot 100 = 66.7\% \text{ by vol.}$$ $$\% \text{butane} = \frac{1}{3} \cdot 100 = 33.3\% \text{ by vol.}$$

Answer: 66.7% by vol. propane and 33.3% by vol. butane (in 3 L mixture: 2 L of propane and 1 L of butane).

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