Question

One litre of co2 is passed through red hot coke. The volume becomes1.4 litre . Find the composition of product. Please a little explanatory....

Answer 1

Answer on Question #44507 – Chemistry – Organic Chemistry

Question

One litre of $\mathrm{CO}_{2}$ is passed through red hot coke. The volume becomes 1.4 litre. Find the composition of product.

Solution

When $\mathrm{CO}_{2}$ is passed through the red hot coke the red-ox reaction occurs, where $\mathrm{CO}_{2}$ acts as an oxidant and the coke (C) acts as a reducer. The reaction results in carbon monoxide (CO) formation:

$$\mathrm{C} + \mathrm{CO}_{2} \rightarrow 2\mathrm{CO}$$

Thus, the product consists of formed CO and unreacted $\mathrm{CO}_{2}$.

The initial volume of $\mathrm{CO}_{2}$:

$$V_{\mathrm{CO}_{2}}^{0} = 1 \text{ litre}$$

The final volume of $\mathrm{CO}_{2}$:

$$V_{\mathrm{CO}_{2}} = x \text{ litre}$$

The volume of CO:

$$V_{\mathrm{CO}} = y \text{ litre}$$

The volume of the product mixture:

$$V_{\text{prod.}} = V_{\mathrm{CO}_{2}} + V_{\mathrm{CO}} = x + y = 1.4 \text{ litre}$$

The volume of $\mathrm{CO}_{2}$ consumed (reacted) equals to the difference between the initial volume and the volume remained after the reaction:

$$V_{\mathrm{CO}_{2}}^{\text{react.}} = V_{\mathrm{CO}_{2}}^{0} - V_{\mathrm{CO}_{2}} = (1 - x) \text{ litre}$$

As is clear from the chemical equation, two moles of CO are formed when 1 mole of $\mathrm{CO}_{2}$ reacts. Volume is proportional to the number of moles, so we can state:

$$V_{\mathrm{CO}} = 2 \cdot V_{\mathrm{CO}_{2}}^{\text{react.}}, \text{ i.e. } y = 2(1 - x)$$

Thus, we have the set of two equations with two unknown values:

$$\left\{ \begin{array}{l} x + y = 1.4 \\ y = 2(1 - x) \end{array} \right.$$

The set can be easily solved by substitution method. Let us substitute the second equation into the first one:

$$\begin{array}{l} x + 2 - 2x = 1.4 \\ x - 2x = 1.4 - 2 \\ x = 0.6 \end{array}$$

Then

$$y = 2 - 2x = 2 - 2 \cdot 0.6 = 2 - 1.2 = 0.8$$

$V_{\mathrm{CO}_{2}} = x = 0.6$ litre, $V_{\mathrm{CO}} = y = 0.8$ litre

So, the 1.4 litre of product mixture consists of 0.6 litre of unreacted $\mathrm{CO}_{2}$ and 0.8 litre of CO.

The product composition in % by vol. (or % by mol.) is as follows

$$\begin{array}{l} \%_{\mathrm{CO}_{2}} = \frac{V_{\mathrm{CO}_{2}}}{V_{\mathrm{prod.}}} 100\% = \frac{0.6 \, l \cdot 100\%}{1.4 \, l} = 42.9\% \text{ by vol.} \\ \%_{\mathrm{CO}} = \frac{V_{\mathrm{CO}}}{V_{\mathrm{prod.}}} 100\% = \frac{0.8 \, l \cdot 100\%}{1.4 \, l} = 57.1\% \text{ by vol.} \\ \end{array}$$

Answer: $\%_{\mathrm{CO}_{2}} = 42.9\%$ by vol., $\%_{\mathrm{CO}} = 57.1\%$ by vol.

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