For the following reaction, 3.93 grams of carbon tetrachloride are mixed with excess methane (CH4). The reaction yields 4.10 grams of dichloromethane (CH2Cl2).
methane (CH4) (g) + carbon tetrachloride (g) dichloromethane (CH2Cl2) (g)
What is the theoretical yield of dichloromethane (CH2Cl2) ? grams
What is the percent yield for this reaction ?
mols CCl4 = grams/molar mass = ?
Convert mols CCl4 to mols of the product using the coefficients in the balanced equation. .
Convert mols of the product to grams. g = mols x molar mass. This is the theoretical yield (TY).
The actual yield (AY) = 4.10 g.
% product = (AY/TY)*100 = ?
To find the theoretical yield of dichloromethane (CH2Cl2), we need to calculate the amount of dichloromethane that can be produced based on the balanced equation and the limiting reactant.
Step 1: Convert the given mass of carbon tetrachloride (CCl4) to moles.
molar mass of CCl4 = 153.82 g/mol
moles of CCl4 = (mass of CCl4) / (molar mass of CCl4)
moles of CCl4 = 3.93 g / 153.82 g/mol
Step 2: Use the stoichiometry of the balanced equation to find the moles of dichloromethane (CH2Cl2) produced from moles of CCl4.
From the balanced equation, we see that the stoichiometric ratio of CCl4 to CH2Cl2 is 1:1.
moles of CH2Cl2 = moles of CCl4
Step 3: Convert moles of CH2Cl2 to grams.
molar mass of CH2Cl2 = 84.93 g/mol
mass of CH2Cl2 = moles of CH2Cl2 * molar mass of CH2Cl2
Now, let's perform the calculations:
Step 1:
moles of CCl4 = 3.93 g / 153.82 g/mol
moles of CCl4 ≈ 0.0255 mol
Step 2:
moles of CH2Cl2 ≈ moles of CCl4
moles of CH2Cl2 ≈ 0.0255 mol
Step 3:
mass of CH2Cl2 ≈ moles of CH2Cl2 * molar mass of CH2Cl2
mass of CH2Cl2 ≈ 0.0255 mol * 84.93 g/mol
mass of CH2Cl2 ≈ 2.17 g
Therefore, the theoretical yield of dichloromethane (CH2Cl2) is approximately 2.17 grams.
To calculate the percent yield, we need to compare the actual yield (given as 4.10 grams) to the theoretical yield.
Step 4: Calculate the percent yield.
percent yield = (actual yield / theoretical yield) * 100%
percent yield = (4.10 g / 2.17 g) * 100%
percent yield ≈ 189.9%
Therefore, the percent yield for this reaction is approximately 189.9%.
To find the theoretical yield of dichloromethane (CH2Cl2), we need to use stoichiometry.
First, let's write and balance the chemical equation for the reaction:
CH4 + CCl4 → CH2Cl2
Next, we need to determine the limiting reactant, which is the reactant that is completely consumed and determines the maximum amount of product that can be formed. To do this, we compare the moles of carbon tetrachloride (CCl4) and methane (CH4) present in their respective amounts given in the problem.
1. We need to convert the given mass of carbon tetrachloride (3.93 grams) to moles. We can use the molar mass of CCl4, which is 153.82 g/mol.
3.93 g CCl4 × (1 mol CCl4 / 153.82 g CCl4) = 0.0255 mol CCl4
2. We need to convert the given mass of dichloromethane (4.10 grams) to moles. We can use the molar mass of CH2Cl2, which is 84.93 g/mol.
4.10 g CH2Cl2 × (1 mol CH2Cl2 / 84.93 g CH2Cl2) = 0.0483 mol CH2Cl2
Now, we compare the moles of CCl4 and CH2Cl2 to determine the limiting reactant:
Moles of CCl4: 0.0255 mol
Moles of CH2Cl2: 0.0483 mol
From the balanced chemical equation, we can see that 1 mole of CCl4 reacts to form 1 mole of CH2Cl2. Therefore, the mole ratio of CCl4 to CH2Cl2 is 1:1.
Since the moles of CH2Cl2 are greater than the moles of CCl4, CCl4 is the limiting reactant. This means that CCl4 will determine the maximum amount of CH2Cl2 that can be formed.
To find the theoretical yield of CH2Cl2, we can use the moles of CCl4 as a starting point:
0.0255 mol CCl4 × (84.93 g CH2Cl2 / 1 mol CH2Cl2) = 2.17 grams CH2Cl2
Therefore, the theoretical yield of dichloromethane (CH2Cl2) is 2.17 grams.
Now, let's calculate the percent yield:
Percent yield = (Actual yield / Theoretical yield) × 100
Given:
Actual yield = 4.10 grams
Theoretical yield = 2.17 grams
Percent yield = (4.10 g / 2.17 g) × 100 = 189.05%
The percent yield is 189.05%.