How much propane did I start with (grams)?

Question

I have 5 grams of water left over from the burning of propane gas. How much did I start with (grams)? What kind of reaction is it?

Answer 1

The burning of propane is a combustion reaction. Assuming a complete combustion (excess oxygen present), then the reaction proceeds by the chemical equation:

C₃H_{8 (g)} + 5O_{2 (g)} --> 3CO_{2 (g)} + 4H₂O _(l)

We are told that at the end of the reaction 5g of water are left over. Since all the Hydrogen comes from the propane, we can use this relation to calculate the amount of propane gas that must have reacted to yield this amount of water. We solve this by using dimensional analysis:

(5g H₂O) x (1 mole H₂O / 18.016 g H₂O) x (1 mole C₃H₈ / 4 mole H₂O) x (44.094 g C₃H₈ / 1 mole C₃H₈)

= 3.059 g C₃H₈

A: We started with approximately 3.059 g of propane. If we use significant figures, then the answer is 3 grams of propane.