Gas formation from reaction #2

Compute the oxygen volume (at standard conditions: 0°C, 1atm) that you need to burn 76.14gr of carbon disulfide in carbon dioxide and sulphur dioxide. Then compute the volume of carbon dioxide and sulphur dioxide and their percentage composition.

SOLUTION:
The balanced reaction that take place is:
$$CS_2(l)+{\color{red} 3}O_2(g) \rightarrow CO_2(g)+{\color{red} 2}SO_2(g)$$

For every mole of $CS_2$ that burns, we need of 3 moles of oxygen. Then:
$$n_{CS_2}=\frac{m_{CS_2}}{MW_{CS_2}}=\frac{76.14gr}{76.14gr/mol}=1mol$$

$$n_{O_2}=3n_{CS_2}$$
$$V_{O_2}=n_{O_2}\cdot22.4=67.2L$$

The volume of carbon dioxide correspond to one mole: 22.4L
The volume of sulpgur dioxide correspond to two moles: 44.8L

The percent composition of $CO_2$ and $SO_2$ is:

$$\%CO_2=\frac{n_{CO_2}}{n_{CO_2}+n_{SO_2}}\cdot100=\frac{1}{1+2}\cdot100=33\%$$

$$\%SO_2=\frac{n_{SO_2}}{n_{CO_2}+n_{SO_2}}\cdot100=\frac{2}{1+2}\cdot100=67\%$$

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PROBLEM 19:
1,00 gram of a mixture of $CaCO_3$ and $MgCO_3$ gives 240mL of $CO_2$ after heating. Compute the percent composition of the mixture (remember that magnesium carbonate and calcium carbonate decompose on heating in the corresponding oxide and carbon dioxide).