For the following reaction, 6.84 grams of benzene (C6H6) are mixed with excess oxygen gas. Assume that the percent yield of carbon monoxide is 71.1 %.
2 C6H6 (l) + 9 O2 (g) 12 CO (g) + 6H2O (g)
(1) What is the theoretical yield of carbon monoxide ? __grams
(2) What is the actual yield of carbon monoxide ? __grams
To find the theoretical yield of carbon monoxide (CO), we need to determine the limiting reactant of the reaction. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's calculate the number of moles of benzene (C6H6) using the given mass of benzene (6.84 grams) and its molar mass (78.11 g/mol):
Number of moles of C6H6 = mass of C6H6 / molar mass of C6H6
= 6.84 g / 78.11 g/mol
Next, we need to calculate the number of moles of CO that can be produced from the calculated moles of benzene. From the balanced equation, we can see that 2 moles of C6H6 produce 12 moles of CO:
Number of moles of CO = (Number of moles of C6H6) x (12 moles of CO / 2 moles of C6H6)
Now, to find the theoretical yield of CO, we need to calculate the mass of CO using its molar mass (28.01 g/mol):
Theoretical yield of CO = Number of moles of CO x molar mass of CO
To find the actual yield of carbon monoxide, we multiply the theoretical yield by the given percent yield:
Actual yield of CO = Theoretical yield of CO x (Percent yield / 100)
By substituting the values into the above equations, you should be able to determine the answers to the questions (1) and (2).
To find the theoretical yield of carbon monoxide (CO), we need to calculate the amount of CO that would be formed if all the benzene reacted completely. We can use stoichiometry to convert the given mass of benzene into moles of CO and then calculate the yield.
Step 1: Calculate the molar mass of benzene (C6H6)
C: 6 x 12.01 g/mol = 72.06 g/mol
H: 6 x 1.01 g/mol = 6.06 g/mol
Total molar mass of benzene = 72.06 + 6.06 = 78.12 g/mol
Step 2: Convert the given mass of benzene to moles
Moles of benzene = 6.84 g / 78.12 g/mol = 0.0876 mol
Step 3: Use the stoichiometry of the reaction to calculate moles of CO
From the balanced equation, the stoichiometric ratio between C6H6 and CO is 2:12, or 1:6.
Moles of CO = 0.0876 mol x 6 = 0.5256 mol
Step 4: Convert moles of CO to grams using the molar mass of CO
Molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Theoretical yield of CO = 0.5256 mol x 28.01 g/mol = 14.7 g
So, the theoretical yield of carbon monoxide is 14.7 grams.
To find the actual yield of carbon monoxide (CO), we need to multiply the theoretical yield by the percent yield.
Step 5: Calculate the actual yield of CO
Actual yield of CO = Theoretical yield of CO x Percent yield / 100
Actual yield of CO = 14.7 g x 71.1 / 100 = 10.4577 g (rounded to 4 decimal places)
Therefore, the actual yield of carbon monoxide is approximately 10.4577 grams.