What mass (g) of water is produced by the combustion of 48.8 g of propane according to the following reaction? Round your answer to one decimal place.

C3H8(g) + 5 O2(g) --> 3 CO2(g) + 4 H2O (l)

To determine the mass of water produced by the combustion of propane, we first need to calculate the moles of propane reacted.

Step 1: Calculate the moles of propane (C3H8).
The molar mass of propane (C3H8) is:
3 carbon atoms (12.01 g/mol) + 8 hydrogen atoms (1.01 g/mol) = 44.11 g/mol

Moles of propane = Mass of propane / Molar mass of propane
Moles of propane = 48.8 g / 44.11 g/mol ≈ 1.107 mol

In the balanced equation, the stoichiometric ratio between propane (C3H8) and water (H2O) is 1:4. This means that for every 1 mole of propane combusted, 4 moles of water are produced.

Step 2: Calculate the moles of water produced.
Moles of water = Moles of propane × (4 mol H2O / 1 mol C3H8)
Moles of water = 1.107 mol × 4 ≈ 4.428 mol

Step 3: Calculate the mass of water produced (H2O).
The molar mass of water (H2O) is:
2 hydrogen atoms (1.01 g/mol each) + 1 oxygen atom (16.00 g/mol) = 18.02 g/mol

Mass of water = Moles of water × Molar mass of water
Mass of water = 4.428 mol × 18.02 g/mol ≈ 79.8 g

Therefore, approximately 79.8 grams of water (H2O) are produced by the combustion of 48.8 grams of propane (C3H8).