How many moles of water can be produced from the reaction of $28 g$ of $C_{3} H_{8}$ ?

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Solution

Combustion of a hydrocarbon gives water.

$C_{3} H_{8_{(g)}} \to 3 C O_{2_{(g)}} + 4 H_{2} O_{(l)}$
We see that $1$ mole of propane gives $3$ moles of carbon dioxide and $4$ moles of water.

Moles of propane $= \frac{28.0 g}{44.1 g m o l - 1} = 0.635 m o l$

And given the stoichiometry of the reaction, the water that arises from the combustion is $0.635 m o l \times 4 = 2.54 m o l$

Therefore, $2.54$ moles of water can be produced.

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