Using C3H8 + 5O2 → 3CO2 + 4H2O How many grams of C3H8 need to be reacted to produce 1.8 L of CO2 at STP?

Molar mass of C₃H₈ = 12.0×3 + 1.0×8 = 44.0 g/mol
Molar volume of CO₂ at STP = 22.4 L/mol

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Mole ratio C₃H₈ : CO₂ = 1 : 3

Moles of CO₂ produced = (1.8 L) / (22.4 L/mol) = 0.08036 mol
Moles of C₃H₈ needed = (0.08036 mol) × (1/3) = 0.02679 mol
Mass of C₃H₈ needed = (0.02679 mol) × (44.0 g/mol) = 1.18 g

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OR:
(1.8 L CO₂) × (1 mol CO₂ / 22.4 L CO₂) × (1 mol C₃H₈ / 3 mol C₃H₈) × (44.0 g C₃H₈ / 1 mol C₃H₈)
= 1.18 g C₃H₈

1.8 L CO2 X (1 mol CO2/22.4 L) X (1 mol C3H8/3 mol CO2) X 44.1 g/mol C3H8 = 1.18 g C3H8