Respuesta :
Answer:
The % yield of the reaction is 16.4 %
Explanation:
Step 1: Data given
Mass of CO2 =2.00 grams
Mass of C5H12 = 4.00 grams
O2 is in excess
Molar mass of CO2 = 44.01 g/mol
Molar mass of C5H12 = 72.15 g/mol
Molar mass of O2 = 32 g/mol
Step 2: The balanced equation
C5H12 + 8O2 → 5CO2 + 6H2O
Step 3: Calculate moles C5H12
Moles C5H12 = mass C5H12 / molar mass C5H12
Moles C5H12 = 4.00 grams / 72.15 g/mol
Moles C5H12 = 0.0554 moles
Step 4: Calculate moles of CO2
For 1 mol pentane, we need 8 moles of O2 to produce 5 moles CO2 and 6 moles H2O
For 0.0554 moles C5H12 we'll have 5*0.0554 = 0.277 moles CO2
Step 5: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.277 moles ¨44.01 g/mol
Mass CO2 = 12.2 grams CO2
Step 6: Calculate % yield
% yield = (actual yield / theoretical yield) *100%
% yield = (2.00 grams / 12.2 grams) * 100%
% yield = 16.4 %
The % yield of the reaction is 16.4 %
The percent yield of the combustion reaction is 16.4 %.
Given that
Mass of CO2 =2.00 grams
Mass of C5H12 = 4.00 grams
The balanced equation
C5H12 + 8O2 → 5CO2 + 6H2O
We first find the No of moles of C5H12 which is given as
No of moles molar mass=Mass/Molar mass
Molar mass of C5H12 = 72.15 g/mol
Moles C5H12 = 4.00 grams / 72.15 g/mol
Moles C5H12 = 0.0554 moles
Then we calculate moles of CO2
Now, 1 mol pentane, 8 moles of O2, 5 moles CO2 and 6 moles H2O
Therefore, 0.0554 moles of C5H12 will give
5 x 0.0554 = 0.277 moles CO2
Mass OF CO2
Soince Molar mass of CO2 = 44.01 g/mol
and nmber of moles of CO2 = 0.277 moles
Then, Mass CO2¨ = 0.277 moles x 44.01 g/mol
Mass CO2 = 12.2 grams CO2
Calculation of Percentage yield= (actual yield / theoretical yield) x 100%
Percentage yield = (2.00 grams / 12.2 grams) x 100%
Percentage yield = 16.4 %
From calculation, Our Percentage yield of the reaction is 16.4 %
See here for more on theoretical yield: https://brainly.com/question/23412624
