calculate the mass of water formed by the complete reaction of 44.1g of propane with oxygen?

When propane reacts with oxygen, the products are carbon dioxide and water

Propane = C3H8
C3H8 + O2 → CO2 + H2O
To balance the C’s, make 3 CO2. To balance the H’s, make 4 H2O.

C3H8 + O2 → 3 CO2 + 4 H2O
To balance the O’s, make 5 O2.
C3H8 + 5 O2 → 3 CO2 + 4 H2O

As you look at the balanced equation, you can see that one mole of propane produces four moles of water.

Mass of one mole = 12 * 3 + 8 = 44 grams
n = 44.1 ÷ 44

For water, n = 176.4 ÷ 44
This is approximately four moles of water. The mass of one mole of water is 18 grams.

Mass = 18 * 176.4 ÷ 44

This is approximately 72 grams of water. I hope this is helpful for you.

Molar mass of C₃H₈ (propane) = (12.01×3 + 1.008×8) g/mol = 44.1 g/mol
Molar mass of H₂O (water) = (1.008×2 + 16.0) g/mol = 18.0 g/mol

Balanced equation for the reaction :
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Mole ratio C₃H₈ : H₂O = 1 : 4

No. of moles of C₃H₈ reacted = (44.1 g) / (44.1 g/mol) = 1.00 mol
No. of moles of H₂O formed = (1.00 mol) × 4 = 4.00 mol
Mass of H₂O formed = (4.00 mol) × (18.0 g/mol) = 72.0 g

C3H8 + 5 O2 → 3 CO2 + 4 H2O

(44.1 g C3H8) / (44.0956 g C3H8/mol) x (4 mol H2O / 1 mol C3H8) x (18.01532 g H2O/mol) = 72.1 g H2O