When 33.3 grams of propane (C3H8) undergoes combustion, what is the theoretical yield of water in grams? The molar mass of p
ropane 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
2 answers:
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 <em>moles of C₃H₈
</em>
Moles C₃H₈ = 33.3 × 1/44
Moles C₃H₈ = 0.757 mol C₃H₈
===============
2. Calculate the <em>moles of H₂O </em>
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 <em>mass of H₂O
</em>
Mass of H₂O = 3.03 × 18
Mass of H₂O = 54 g
C3H8+ 5 O2 --> 3 CO2 + 4 H2O
44 g. --------> 72 g
33.3 g. --------> x

Answer: The theoretical yield of H2O is 54.5
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