Question

Copper(II) carbonate is heated in a test tube.Gas produced is passed in lime water through a delivery tube.6.2g of copper(ii) carbonate is used in the reaction.Calculate the volume of carbon dioxide gas produced at room condition.

Answer 1

Answer on Question #48733 - Chemistry – Inorganic Chemistry

Question

Copper(II) carbonate is heated in a test tube. Gas produced is passed in lime water through a delivery tube. 6.2g of copper(ii) carbonate is used in the reaction. Calculate the volume of carbon dioxide gas produced at room condition.

Answer:

Heating of copper(II) carbonate:

$$\mathrm{CuCO_3(s)} = \mathrm{CuO(s)} + \mathrm{CO_2(g)}$$

$\mathrm{CO_2}$ is a produced gas.

Number of moles of copper(II) carbonate equals:

$$n(\mathrm{CuCO_3}) = \frac{m(\mathrm{CuCO_3})}{M(\mathrm{CuCO_3})} = \frac{6.2}{123.55} = 0.05 \text{ moles}$$

According to the reaction equation, number of moles of $\mathrm{CO_2}$ produced is equal to that of copper(II) carbonate:

$$n(\mathrm{CO_2}) = n(\mathrm{CuCO_3}) = 0.05 \text{ moles}$$

Then the volume of carbon dioxide $(\mathrm{CO_2})$ is:

$$V(\mathrm{CO_2}) = n(\mathrm{CO_2}) \cdot V_m = 0.05 \cdot 22.4 = 1.12 \text{ L}$$

Answer: 1.12 L of $\mathrm{CO_2}$

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