When 33.3 grams of propane (c3h8) undergoes combustion, what is the theoretical yield of water in grams? the molar mass of propane is 44 g/mol and the molar mass of water is 18 g/mol. input your answer without the units. c3h8+ 5 o2 --> 3 co2 + 4 h2o

54 g

Explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

M_r:        44                                   18

             C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

Mass/g: 33.3

1. Calculate the moles of C₃H₈

Moles C₃H₈ = 33.3 × 1/44

Moles C₃H₈ = 0.757 mol C₃H₈

2. Calculate the moles of H₂O  

The molar ratio is 4 mol H₂O:1 mol C₃H₈.

Moles of NaCl = 0.757 × 4/1

Moles of NaCl = 3.03 mol H₂O

3. Calculate the mass of H₂O

Mass of H₂O = 3.03 × 18

Mass of H₂O = 54 g

the answer is b & c i believe : )