Question

What is the volume of $C{O}_2$ gas obtained on the complete combustion of $30\text{mL}$ of propane gas?

• A. $10\text{mL}$
• B. $30\text{mL}$
• C. $90\text{mL}$
• D. $40\text{mL}$

Answer 1

The correct option is C $90\text{mL}$

Combustion of propane:
$C_3{H}_8\text{(g)}+5O_2\text{(g)}\to 3C{O}_2\text{(g)}+4H_{2}O\text{(l)}$
Gay-Lussac’s law of combining volume states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure conditions remain constant. In the above equation:
1 volume of propane yields 3 volumes of $C{O}_2$
So, 30 mL propane yields $=3\times 30\text{mL}$ of $C{O}_2=90\text{mL}C{O}_2$