If 156.06 g of propane, C3H8, is burned in excess oxygen, how many grams of water are formed? C3H8 + O2 → CO2 + H2O Select one: a. 255.0 b. 765.1 c. 191.3 d. 31.88

The correct option is;

a. 255.0 g

Explanation:

The given information are;

Mass of propane, C₃H₈ in the combustion reaction = 156.06 g

The equation of the combustion reaction is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

From the balanced chemical equation of the reaction, we have;

One mole of propane, C₃H₈ reacts with five moles oxygen gas, O₂, to form three moles of carbon dioxide, CO₂, and  four moles of water, H₂O

The molar mass of propane gas = 44.1 g/mol

The number of moles, n, of propane gas = Mass of propane gas/(Molar mass of propane gas) = 156.06/44.1 = 3.54 moles

Given that one mole of propane gas produces 4 moles of water molecule (steam) H₂O, 3.54 moles of propane gas will produce 4×3.54 = 14.16 moles of  (steam) H₂O

The mass of one mole of H₂O = 18.01528 g/mol

The mass of 14.16 moles of H₂O = 14.16 × 18.01528 = 255.0 g

The mass of H₂O produced = 255.0 g