Answer:
[tex]\huge\boxed{\sf 39\ L}[/tex]
Explanation:
In the reaction:
for 1 mol of propane, 3 moles of CO₂ are produced.
Volume of propane = 13 L
At STP,
1 mol of propane = 22.4 L/dm³ molar value
So,
Moles in 13 L of propane:
= volume / molar volume
= 13 / 22.4
= 0.58 mol
Since,
1 mol of propane = 3 moles of CO₂
0.58 mol of propane = 3 × 0.58 mol of CO₂
0.58 mol of propane = 1.74 mol of CO₂
Converting 1.74 mol of CO₂ into L:
[Multiply it by 22.4 L/mol]
= 1.74 mol × 22.4 L/mol
= 39 L
[tex]\rule[225]{225}{2}[/tex]
