Respuesta :
Answer:
84.86%
Explanation:
Step 1:
We'll begin by writing a balanced equation for the reaction between C8H18 and O2 to produce CO2. This is illustrated below:
2C8H18 + 25O2 —> 16CO2 + 18H2O
Step 2:
Now, let us calculate the mass of O2 that reacted and the mass of CO2 produced from the balanced equation above. This is illustrated below:
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 25 x 32 = 800g
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 16 x 44 = 704g
Therefore the mass of O2 that reacted from the balanced equation is 800g
The mass of CO2 produced from the balanced equation is 704g
Step 3:
Determination of the theoretical yield of CO2. This is illustrated below:
From the balanced equation above,
800g of O2 reacted to produced 704g of CO2.
Therefore, 218g of O2 will react to produce = (218 x 704)/800 = 191.84g of CO2.
Therefore, the theoretical yield of CO2 is 191.84g
Step 4:
Determination of the percentage yield of CO2. This is illustrated below:
Actual yield = 162.8g
Theoretical yield = 191.84g
Percentage yield =?
Percentage yield = Actual yieldm/Theoretical yield x100
Percentage yield = 162.8/191.84 x100
Percentage yield = 84.86%
Therefore, the percentage yield of CO2 is 84.86%
Answer:
The percent yield of the reactions is 84.8 %
Explanation:
Step 1: data given
Mass of CO2 formed = 162.8 grams
Mass of O2 = 218.0 grams
C8H18 is in excess
Molar mass O2 = 32.0 g/mol
Molar mass of CO2 = 44.01 g/mol
Step 2: The balanced equation
2C8H18 + 25O2 → 16CO2 + 18H2O
Step 3: Calculate moles O2
Moles O2 = 218.0 grams / 32.0 g/mol
Moles O2 = 6.8125 moles
Step 4: Calculate moles CO2
For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O
For 6.8125 moles O2 we'll have 16/25 * 6.8125 = 4.36 moles CO2
Step 5: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 4.36 moles ¨¨ 44.01 g/mol
Mass CO2 = 191.9 grams
Step 6: Calculate the percent yield
Percent yield = (actual mass / theoretical mass) * 100 %
Percent yield = (162.8 grams / 191.9 grams ) * 100 %
Percent yield = 84.8 %
The percent yield of the reactions is 84.8 %
