# calculate the volume of oxygen required for the complete combustion of 8.8 g of propane

Solution-

The balanced equation of Combustion of propane is-

C3H8 + 5O2  3CO2 + 4H2O

The molecular mass of propane = 44 gm
The volume of  5 O2 at STP = 5 $×$ 22.4 = 112 L
So, 44g of propane requires 112 L of oxygen for complete combustion.

8.8 gm of propane requires oxygen for combustion = $\frac{112}{44}×8.8$ = 22.4 L

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