Question

The combustion of propane, C₃H₈(g), proceeds according to the equation:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) In the complete combustion of 44.0 grams of propane, how many grams of water will be formed?

a) 18.02 g
b) 36.04 g
c) 54.06 g
d) 72.08 g

Answer 1

To find the grams of water formed in the complete combustion of propane, we can use the balanced chemical equation: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l). From the equation, we can see that 1 mole of propane produces 4 moles of water. Therefore, the number of grams of water formed is 72.08 g.

Step-by-step explanation:

To find the grams of water formed in the complete combustion of propane, we can use the balanced chemical equation:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

From the equation, we can see that 1 mole of propane produces 4 moles of water. To find the moles of propane, we divide the given mass (44.0 g) by the molar mass of propane (44.1 g/mol). This gives us 1 mole of propane. Therefore, 1 mole of propane will produce 4 moles of water.

The molar mass of water is 18.02 g/mol. Therefore, the number of grams of water formed will be 4 moles x 18.02 g/mol = 72.08 g.

Therefore, the correct answer is d) 72.08 g.