Question

At \[25^\circ C\] for complete combustion of \[5mol\] propane (\[{C_3}{H_8}\]). The required volume of \[{O_2}\] at STP will be? For \[{C_3}{H_8}\], the combustion reaction is

Answer 1

Hint: Combustion reaction refers to the complete burning of a substance in presence of oxygen. Propane is a hydrocarbon in the alkane family with three carbon atoms.

Complete step by step answer:
-Mole of a substance or compound is defined as the ratio of the amount of the substance taking part in a reaction and the molecular weight or molar mass of the compound.
-Thus mole of reactant $ = \dfrac{\text{Amount of reactant}}{\text{molar mass of reactant}}$
-The given reaction is the combustion of propane. Propane is a hydrocarbon which is composed of carbon and hydrogen. It is also known as alkane with three carbon atoms.
-The combustion of propane with oxygen produces carbon dioxide and water. The balanced equation for the combustion reaction of propane is
${C_3}{H_8} + 5{O_2} \to 3C{O_2} + 4{H_2}O$
-From the above reaction it is evident that one equivalent of propane requires five equivalents of oxygen to generate three equivalents of carbon dioxide and four equivalents of water.
-At STP means under standard temperature and pressure the molar volume of a gas is \[22.4litres\] . In other words at 0oC temperature and 1atm pressure the one mole of a gas occupies a volume of \[22.4litres\].
-Thus one mole of \[{O_2}\] = \[22.4litres\].
-For the desired combustion reaction one mole of propane requires five mole of oxygen. Thus one mole of propane requires $5 \times 22.4 = 112litres$ of oxygen.
-For complete combustion of \[5mol\] of propane, the volume of oxygen required is $5 \times 112 = 560litres$.

Note:
Oxygen serves as fuel for the combustion reaction. Combustion of a substance produces heat and light energy as products. Complete combustion of substance generally produces blue flame.