For the complete combustion of propane, given by C3H8+5O2 →3CO2+4H2O, what mass of water is produced when 13.1 g of propane is burned in air?

Answer 1

Step 1: Convert mass of propane to moles of propane.

The molar mass of propane is 44.094 g/mol.

13.1 g C₃H₈ x (1 mol C₃H₈ / 44.094 g C₃H₈) = 0.297 mol C₃H₈

Step 2: Use the balanced equation to determine how many moles of H₂O are produced.

From the balanced equation, we can see that 1 mole of C₃H₈ produces 4 moles of water.

0.297 mol C₃H₈ x (4 mol H₂O / 1 mol C₃H₈) = 1.188 mol H₂O

Step 3: Convert moles of H₂O to grams of H₂O using the molar mass of H₂O (18.016 g/mol).

Using the molar mass, convert from moles of H₂O to grams of H₂O.

1.188 mol H₂O x (18.016 g H₂O / 1 mol H₂O) = 21.41 g H₂O

Answer: Complete combustion of 13.1 g of propane would produce 21.4 g of H₂O.

You can also do all of these conversions in one calculation using dimensional analysis.

13.1 g C₃H₈ x (1 mol C₃H₈ / 44.094 g C₃H₈) x (4 mol H₂O / 1 mol C₃H₈) x (18.016 g H₂O / 1 mol H₂O) = 21.4 g H₂O.

Hope this helps!