Question

Suppose that an excess of propane, C₃H₈, burns in 320 g of O₂. How many moles of H₂O will be formed?

1) 2 moles H₂O
2) 4 moles H₂O
3) 6 moles H₂O
4) 8 moles H₂O

Answer 1

When an excess of propane burns in 320 g of O₂, 8 moles of H₂O will be formed after converting the mass of O₂ to moles and using the stoichiometry of the complete combustion of propane.

Step-by-step explanation:

To calculate the number of moles of H₂O formed when propane (C₃H₈) burns in 320 g of O₂, we need the balanced equation for the complete combustion of propane:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

According to the balanced equation, 5 moles of O₂ produce 4 moles of H₂O. First, we convert the mass of O₂ to moles using the molar mass of O₂, which is approximately 32 g/mol:

• 320 g O₂ × (1 mol O₂ / 32 g O₂) = 10 moles O₂

Next, we use the stoichiometry of the reaction to find the moles of H₂O produced:

• (10 moles O₂) × (4 moles H₂O / 5 moles O₂) = 8 moles H₂O

Therefore, 8 moles of H₂O will be formed when an excess of propane burns in 320 g of O₂.