Question

When 33.3 grams of propane (C​3​H​8​) 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

Answer 1

Answer:

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

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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

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3. Calculate the mass of H₂O

Mass of H₂O = 3.03 × 18

Mass of H₂O = 54 g

Answer 2

C3H8+ 5 O2 --> 3 CO2 + 4 H2O
44 g. --------> 72 g
33.3 g. --------> x
$x = \frac{33.3 \times 72}{44} \\ x = 54.5 \: g$
Answer: The theoretical yield of H2O is 54.5