Calculate the mass of water produced when 42 g of propane, c3h8, is burned with 115 g of oxygen

The balanced combustion reaction of propane, C₃H₈, is

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Molar mass of propane: 44 g/mol
Moles of propane = 42 g * (1 mol/44g) = 0.9545 mol propane

Molar mass of oxygen: 32 g/mol
Moles of oxygen = 115 g * (1 mol/32 g) = 3.594 mol oxygen

Moles of oxygen needed to completely react propane:
0.9545 mol propane * (5 mol O₂/1 mol propane) = 4.7725 mol oxygen

Since the available oxygen is only 3.594 moles and propane needs 4.7725 moles, that means oxygen is our limiting reactant. We base the amount of water produced here.

Molar mass of water: 18 g/mol
Mass of water produced = 3.594 mol O₂ * (4 mol H₂O/5 mol O₂) * (18 g/mol)
Mass of water produced = 258.768 grams

The answer is a) carbon-12