Answer to Question #116700 in Chemistry for lin

Question

If 500g of marble are decomposed completely and knowing that it contains 90% calcium carbonate, calculate the mass of calcium oxide and the volume of carbon dioxide ( at S.T.P.) produced.

Answer 1

The decomposition of calcium carbonate consists in a following reaction:

CaCO₃ $\rightarrow$ CaO + CO₂$\uparrow$ .

According to this reaction, the number of the moles of calcium oxide, carbon oxide and calcium carbonate relate as:

$n(CaCO_3)=n(CO_2)=n(CaO)$ .

The number of the moles of calcium oxide is its mass (500 g · 0.9 = 450 g) divided by its molar mass (100.09 g/mol):

$n(CaCO_3) = \frac{450 \text{ g}}{100.09 \text{ g/mol}} = 4.50$ mol.

Therefore, the number of the moles of CO₂ and of CaO are:

$n(CaO)=n(CO_2) = 4.50$ mol.

The mass of calcium oxide produced is a product of its number of the moles and its molar mass (56.08 g/mol):

$n(CaO)=4.50·56.08 = 252$ g.

The volume of carbon dioxide produced can be calculated from the ideal gas law:

$V(CO_2) = \frac{nRT}{p} = \frac{4.50\text{ mol}·8.314 \text{ m}^3\text{Pa K}^{-1} \text{mol}^{-1}·273.15\text{ K}}{10^5\text{ Pa}}$

$V(CO_2) = 0.102$ m³.

Answer: If 500g of marble are decomposed completely and knowing that it contains 90% calcium carbonate, 252 g of calcium oxide and 0.102 m³ of carbon dioxide ( at S.T.P.) produced.