In a reaction between methane, CH4+ and chlorine, Cl2+ four products can form: CH3Cl, CH2Cl2 + CHCl3+ and CCl4. In a particular instance, 20.8 g of CH4 were allowed to react with excess Cl2 and gave 5.0 g CH3Cl, 25.5 g CH2Cl2+ and 59.0 g CHCl3+ all the CH4 reacted.
a) How many grams of CCl4 were formed?
b) On the basis of available CH4+ what is the theoretical yield of CCl4?
c) What is the percentage yield of CCl4?
d) How many grams of Cl4 reacted with the CH4?
See your other post under Sunny.
a) To determine the grams of CCl4 formed, we need to find the mass of CCl4 produced.
Given:
Mass of CH4 = 20.8 g
Mass of CH3Cl formed = 5.0 g
Mass of CH2Cl2 formed = 25.5 g
Mass of CHCl3 formed = 59.0 g
The chemical equation for the reaction is:
CH4 + Cl2 → CH3Cl + CH2Cl2 + CHCl3 + CCl4
We know that the total mass of the products formed is equal to the mass of the reactant(CH4) used.
Total mass of products formed = Mass of CH3Cl + Mass of CH2Cl2 + Mass of CHCl3 + Mass of CCl4
= 5.0 g + 25.5 g + 59.0 g + Mass of CCl4
Since all the CH4 reacted, the total mass of the products formed is equal to the mass of CH4 used.
Total mass of products formed = Mass of CH4
= 20.8 g
Therefore, we can write the equation as:
20.8 g = 5.0 g + 25.5 g + 59.0 g + Mass of CCl4
Simplifying the equation:
Mass of CCl4 = 20.8 g - 5.0 g - 25.5 g - 59.0 g
= -68.7 g
The calculated mass of CCl4 is negative, which indicates an error in the calculation or inconsistent data. Please double-check the given values and calculations.
b) To determine the theoretical yield of CCl4 on the basis of available CH4:
The theoretical yield is the maximum amount of product that can be formed based on stoichiometry.
Looking at the balanced chemical equation:
1 mol CH4 produces 1 mol CCl4
The molar mass of CH4 = 16.04 g/mol
The molar mass of CCl4 = 153.82 g/mol
Now, let's convert the given mass of CH4 into moles:
Moles of CH4 = Mass of CH4 / Molar mass of CH4
= 20.8 g / 16.04 g/mol
= 1.297 mol
According to the stoichiometry, 1 mol of CH4 produces 1 mol of CCl4. Therefore, the theoretical yield of CCl4 is equal to the moles of CH4 used.
Theoretical yield of CCl4 = Moles of CH4
= 1.297 mol
c) The percentage yield of CCl4 can be calculated by dividing the actual yield (mass of CCl4 formed) by the theoretical yield and multiplying by 100.
Percentage yield of CCl4 = (Actual yield / Theoretical yield) x 100
However, since we obtained a negative value for the mass of CCl4 in part a, we cannot calculate the percentage yield. Please double-check the given values and calculations.
d) To calculate the grams of Cl2 reacted with CH4:
Since all the CH4 reacted and the reaction stoichiometry is 1:1 between CH4 and Cl2,
Grams of Cl2 reacted = Grams of CH4 used
= 20.8 g
Therefore, 20.8 grams of Cl2 reacted with CH4.
To solve these questions, we need to use stoichiometry, which is the study of the quantitative relationships between reactants and products in a chemical reaction. Stoichiometry allows us to calculate the amount of products formed based on the amounts of reactants.
First, let's write the balanced chemical equation for the reaction:
CH4 + Cl2 -> CH3Cl + CH2Cl2 + CHCl3 + CCl4
a) To find the grams of CCl4 formed, we need to determine the molar mass of CH3Cl, CH2Cl2, CHCl3, and CCl4.
- CH3Cl (methyl chloride) has a molar mass of 50.49 g/mol
- CH2Cl2 (dichloromethane) has a molar mass of 84.93 g/mol
- CHCl3 (trichloromethane) has a molar mass of 119.38 g/mol
- CCl4 (carbon tetrachloride) has a molar mass of 153.82 g/mol
Next, we can use the given masses of CH3Cl, CH2Cl2, and CHCl3 to find the number of moles for each product.
- Moles of CH3Cl = mass of CH3Cl / molar mass of CH3Cl
- Moles of CH2Cl2 = mass of CH2Cl2 / molar mass of CH2Cl2
- Moles of CHCl3 = mass of CHCl3 / molar mass of CHCl3
Now, let's calculate these values using the given masses:
Moles of CH3Cl = 5.0 g / 50.49 g/mol = 0.099 mol
Moles of CH2Cl2 = 25.5 g / 84.93 g/mol = 0.300 mol
Moles of CHCl3 = 59.0 g / 119.38 g/mol = 0.494 mol
Since all the CH4 reacted, we can assume that the moles of CH4 is equal to the sum of the moles of the products.
Total moles of products = Moles of CH3Cl + Moles of CH2Cl2 + Moles of CHCl3 + Moles of CCl4
Total moles of products = 0.099 mol + 0.300 mol + 0.494 mol + Moles of CCl4
Since we are asked to find the grams of CCl4, we need to convert the moles of CCl4 to grams using its molar mass of 153.82 g/mol.
- Moles of CCl4 = Total moles of products - (Moles of CH3Cl + Moles of CH2Cl2 + Moles of CHCl3)
- Moles of CCl4 = 0.893 mol - (0.099 mol + 0.300 mol + 0.494 mol)
- Moles of CCl4 = 0.893 mol - 0.893 mol
- Moles of CCl4 = 0
Since there are no moles of CCl4, the mass of CCl4 formed is zero.
Therefore, a) The amount of CCl4 formed is 0 grams.
b) The theoretical yield of CCl4 is the maximum amount of CCl4 that can be produced based on the amount of available CH4. Since all the CH4 reacted and no CCl4 formed, the theoretical yield of CCl4 is also 0 grams.
c) The percentage yield of CCl4 is a measure of the efficiency of the reaction. It is calculated by dividing the actual yield (0 grams) by the theoretical yield (0 grams) and multiplying by 100.
Percentage yield = (Actual yield / Theoretical yield) * 100
Percentage yield = (0 g / 0 g) * 100
Percentage yield is undefined.
d) Since all the CH4 reacted, the moles of Cl2 needed was in a 1:1 ratio with the moles of CH4. Therefore, the moles of Cl2 consumed is equal to the moles of CH4.
Moles of Cl2 reacted = Moles of CH4 = 0.893 mol
To find the grams of Cl2 reacted, we need to use its molar mass of 70.91 g/mol.
Mass of Cl2 reacted = Moles of Cl2 reacted x Molar mass of Cl2
Mass of Cl2 reacted = 0.893 mol x 70.91 g/mol
Mass of Cl2 reacted = 63.44 g
Therefore, d) The mass of Cl2 reacted with the CH4 is 63.44 grams.