Answer:
81.3%
Explanation:
Step 1:
The balanced equation for the reaction:
This is shown below:
C3H8 + 5O2 —> 3CO2 + 4H2O
Step 2:
Data obtained from the question. This includes:
Mass of propane (C3H8) = 470 g
Actual yield of water (H2O) = 625 g
Percentage yield of water (H2O) =?
Step 3:
Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:
C3H8 + 5O2 —> 3CO2 + 4H2O
Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol
Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol
Mass of H2O from the balanced equation = 4 x 18 = 72g
From the balanced equation above,
44g of C3H8 was burned and 72g of H2O was produced.
Step 4:
Determination of the theoretical yield of H2O. This is illustrated below:
From the balanced equation above,
44g of C3H8 produced 72g of H2O.
Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.
Therefore, the theoretical yield of H2O is 769.09g
Step 5:
Determination of the percentage yield of water (H2O). This is illustrated below:
Actual yield of water (H2O) = 625g
theoretical yield of H2O = 769.09g
Percentage yield of water (H2O) =?
Percentage yield = Actual yield/Theoretical yield x100
Percentage yield = 625/769.09 x100
Percentage yield = 81.3%
Therefore, the percentage yield of water (H2O) is 81.3%
= 50 g / mole