how many grams of oxygen gas was required for the reaction to occur?

Question

The combustion of propane gas is used to fuel barbecue grills. If 57.3 grams of water vapor were produced from a grilling session, how many grams of oxygen gas was required for the reaction to occur?

1 C₃H₈ (g) + 5 O₂ (g) → 3 CO₂ (g) + 4 H₂O (l)

Answer 1

Hello, Rosie,

Thanks for providing an already-balanced equation. The way to answer the question relies on the use of moles in the calculation. The equation tells us that it takes 5 moles of O₂ to produce 4 moles of H₂O. If we calculate the number of moles of CO₂ produced, we can multiply that ratio (5 moles O₂/4 moles H₂O) to find moles O₂ consumed.

Divide 57.3 grams H₂O by it's molar mass, 18 g/mole. That yields 3.18 moles H₂O.

(3.18 moles H₂O)*(5 moles O₂/4 moles H₂O) = 3.98 moles O₂.

Multiply moles O₂ by it's molar mass to get grams O₂.

(3.98 moles O₂)*(32 g/mole) = 127 grams O₂

I hope this helps,

Bob