Answer:
Percentage yield of H₂O = 74.24%
Step-by-step explanation:
The balanced equation for the reaction is given below:
2C₆H₁₀ + 17O₂ —> 12CO₂ + 10H₂O
Next, we shall determine the masses of C₆H₁₀ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This is can be obtained as follow:
Molar mass of C₆H₁₀ = 82 g/mol
Mass of C₆H₁₀ from the balanced equation = 2 × 82 = 164 g
Molar mass of O₂ = 32 g/mol
Mass of O₂ from the balanced equation = 17 × 32 = 544 g
Molar mass of H₂O = 18 g/mol
Mass of H₂O from the balanced equation = 10 × 18 = 180 g
SUMMARY:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂ to produce 180 g of H₂O.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂.
Therefore, 115 g of C₆H₁₀ will react to produce = (115 × 544)/164 = 381 g of O₂.
From the calculations made above, we can see that a higher mass (i.e 381 g) of O₂ than what was given (i.e 199 g) is needed to react with 115 g of C₆H₁₀.
Therefore, O₂ is the limiting reactant and C₆H₁₀ is the excess reactant.
Next, we shall determine the theoretical yield of H₂O. This can be obtained by using the limiting reactant as shown below:
From the balanced equation above,
544 g of O₂ reacted to produce 180 g of H₂O.
Therefore, 199 g of O₂ will react to produce = (199 × 180)/544 = 66 g of H₂O.
Thus, the theoretical yield of H₂O is 66 g.
Finally, we shall determine the percentage yield. This can be obtained as follow:
Actual yield of H₂O = 49 g
Theoretical yield of H₂O = 66 g
Percentage yield of H₂O =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield of H₂O = 49/66 × 100
Percentage yield of H₂O = 74.24%