The volume of butane that can be burned in oxygen with a volume of 65 liters is …

First, we compose the equation of reactions, then we equalize it.
2С4Н10 + 13О2 = 8СО2 + 10Н2О.
Further, according to the reaction equation, we find the volume of butane. We write down the solution.
Chl С4Н10 – 65 l О2.
2 × 22.4 L / mol – 13 × 22.4 L / mol O2.
Find the unknown value of x.
X = 2 × 22.4 × 65 ÷ (13 × 22.4) = 10 liters of butane.
This means that in oxygen with a volume of 65 liters, 10 liters of butane can be burned.
Answer: the volume of butane is 10 liters.