How many litres of oxygen at S.T.P. will be required to burn completely 2.2 g of propane?

Solution

The balanced chemical equation representing the reaction is
C₃H₈ + 5O₂

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3CO₂ + 4H₂O
1 mole 5 x 22.4L
44 g at STP
Stoichiometry indicates that
1 mole (44 g) propane

identical to space 5 space cross times space 22.4 space straight L space space space straight O subscript 2 space at space STP

therefore space space 2.2 space straight g space propane space equals space fraction numerator 5 space cross times space 22.4 space cross times space 2.2 over denominator 44 end fraction litre space straight O subscript 2 space space space space space space space space space space space space space space space space space space space space space space space space space equals 5.6 space litre space straight O subscript 2 space at space STP

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