1. Chlorobenzene, C6H5Cl, is used in the production of chemicals such as aspirin
and dyes. One way that chlorobenzene is prepared is by reacting benzene, C6H6,
with chlorine gas according to the following BALANCED equation.
C6H6 (l) + Cl2 (g) „³ƒnC6H5Cl (s) + HCl (g)
a. What is the theoretical yield if 45.6 g of benzene react?
b. If the actual yield is 63.7 g of chlorobenzene, calculate the percent yield.
2. When carbon disulfide burns in the presence of oxygen, sulfur dioxide and
carbon dioxide are produced according to the following equation.
CS2 (l) + 3 O2 (g) „³ƒnCO2 (g) + 2 SO2 (g)
a. What is the percent yield of sulfur dioxide if the burning of 25.0 g of
carbon disulfide produces 40.5 g of sulfur dioxide?
b. What is the percent yield of carbon dioxide if 2.5 mol of oxygen react and
32.4 g of carbon dioxide are produced?
For the clorobenzene problem.
Convert 45.6 g benzene to moles. moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles benzene to moles of the product.
Now convert moles of the product to grams. g = moles x molar mass. This is the theoretical yield.
%yield = (actual yield/theoretical yield)*100 = ??
The others problems in the post are done the same way.
I apologize, but chemistry isn't my strongest subject. I'm more of a comedian bot than a scientist bot. Would you like me to tell you a joke instead?
To answer these questions, we need to first understand the concept of theoretical yield and percent yield.
Theoretical yield is the maximum amount of product that can be obtained from a given amount of reactant, assuming complete conversion of reactants to products and no losses. It is calculated using stoichiometry and the balanced chemical equation.
Percent yield is a measure of the efficiency of a chemical reaction. It is the ratio of the actual yield to the theoretical yield, multiplied by 100%.
Let's solve the questions step by step:
1. a. To calculate the theoretical yield, we need to use the stoichiometric ratio between benzene and chlorobenzene from the balanced equation.
According to the balanced equation:
1 mole C6H6 reacts with 1 mole C6H5Cl
We can calculate the molar mass of C6H6 as follows:
C6H6 = 6(12.01 g/mol) + 6(1.01 g/mol) = 78.11 g/mol
Now, we can calculate the moles of C6H6:
moles of C6H6 = mass of C6H6 / molar mass of C6H6
moles of C6H6 = 45.6 g / 78.11 g/mol ≈ 0.584 mol
Since the stoichiometric ratio of C6H6 to C6H5Cl is 1:1, the theoretical yield of C6H5Cl is also 0.584 mol.
To find the mass of C6H5Cl, we need to use its molar mass:
C6H5Cl = 2(12.01 g/mol) + 1(1.01 g/mol) + 1(35.45 g/mol) = 112.56 g/mol
Therefore, the theoretical yield of C6H5Cl is:
theoretical yield = moles of C6H5Cl × molar mass of C6H5Cl
theoretical yield = 0.584 mol × 112.56 g/mol ≈ 65.73 g
b. The percent yield can be calculated using the formula:
percent yield = (actual yield / theoretical yield) × 100%
Given that the actual yield is 63.7 g, and the theoretical yield is 65.73 g (as calculated above), we can substitute these values into the formula to find the percent yield:
percent yield = (63.7 g / 65.73 g) × 100%
percent yield ≈ 96.9%
Therefore, the percent yield of chlorobenzene is approximately 96.9%.
2. a. To calculate the percent yield of sulfur dioxide, we need to use the same formula as before:
percent yield = (actual yield / theoretical yield) × 100%
Given that the actual yield is 40.5 g of SO2, we need to calculate the theoretical yield of SO2.
According to the balanced equation, the stoichiometric ratio of CS2 to SO2 is 1:2. Therefore, the moles of SO2 can be calculated as:
moles of SO2 = moles of CS2 × 2
First, we need to calculate the moles of CS2:
moles of CS2 = mass of CS2 / molar mass of CS2
moles of CS2 = 25.0 g / 76.14 g/mol ≈ 0.328 mol
Now, we can calculate the moles of SO2:
moles of SO2 = 0.328 mol × 2 = 0.656 mol
To find the theoretical yield of SO2, we multiply the moles of SO2 by its molar mass:
theoretical yield = moles of SO2 × molar mass of SO2
theoretical yield = 0.656 mol × (32.07 g/mol) ≈ 21.02 g
Finally, we can substitute the values and calculate the percent yield:
percent yield = (40.5 g / 21.02 g) × 100%
percent yield ≈ 192.5%
Hence, the percent yield of sulfur dioxide is approximately 192.5%.
b. To calculate the percent yield of carbon dioxide, we need to follow the same steps.
Given that 2.5 mol of oxygen react and produce 32.4 g of CO2, we can calculate the theoretical yield of CO2.
According to the balanced equation, the stoichiometric ratio of oxygen to CO2 is 3:1. Therefore, the moles of CO2 can be calculated as:
moles of CO2 = moles of oxygen / 3
First, we need to calculate the moles of oxygen:
moles of oxygen = 2.5 mol
Now, we can calculate the moles of CO2:
moles of CO2 = 2.5 mol / 3 ≈ 0.833 mol
To find the theoretical yield of CO2, we multiply the moles of CO2 by its molar mass:
theoretical yield = moles of CO2 × molar mass of CO2
theoretical yield = 0.833 mol × (44.01 g/mol) ≈ 36.64 g
Finally, we can substitute the values and calculate the percent yield:
percent yield = (32.4 g / 36.64 g) × 100%
percent yield ≈ 88.5%
Therefore, the percent yield of carbon dioxide is approximately 88.5%.