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?
this is useless
Find tge product formed in these
1.CH4 + CL2=
2. C3H8 + O2
To answer these questions, we need to use stoichiometry, which involves the balanced equation and the molar masses of the compounds involved.
a) To find the grams of CCl4 formed:
From the balanced equation, we know that:
1 mole of CH4 reacts with 1 mole of CCl4.
We are given the mass of CH4 formed (20.8 g) and the molecular mass of CH4 (16.04 g/mol). We can use these values to calculate the number of moles of CH4:
moles of CH4 = mass of CH4 / molecular mass of CH4
= 20.8 g / 16.04 g/mol
= 1.296 mol
Since the ratio of CH4 to CCl4 is 1:1, we can conclude that 1.296 mol of CH4 reacts to form 1.296 mol of CCl4.
To find the mass of CCl4:
mass of CCl4 = moles of CCl4 x molecular mass of CCl4
= 1.296 mol x (12.01 g/mol + 4 x 35.45 g/mol)
= 1.296 mol x 153.82 g/mol
= 199.27 g
Therefore, 199.27 grams of CCl4 were formed.
b) To find the theoretical yield of CCl4:
Theoretical yield refers to the maximum amount of product that can be obtained assuming the reaction goes to completion. In this case, it is calculated based on the limiting reactant, which is the reactant that is completely consumed during the reaction.
From the stoichiometry, we know that 1 mole of CH4 reacts with 1 mole of CCl4. Therefore, the number of moles of CCl4 formed will be equal to the moles of CH4.
moles of CCl4 = moles of CH4
= 1.296 mol
Using the molecular mass of CCl4 (153.82 g/mol), we can calculate the theoretical yield:
theoretical yield of CCl4 = moles of CCl4 x molecular mass of CCl4
= 1.296 mol x 153.82 g/mol
= 199.27 g
The theoretical yield of CCl4 is 199.27 grams.
c) To find the percentage yield of CCl4:
Percentage yield is calculated by comparing the actual yield (obtained in the experiment) with the theoretical yield.
Actual yield: 59.0 g (given)
Percentage yield = (Actual yield / Theoretical yield) x 100
= (59.0 g / 199.27 g) x 100
= 29.6%
The percentage yield of CCl4 is 29.6%.
d) To find the grams of Cl2 reacted with CH4:
From the balanced equation, we know that 1 mole of CH4 reacts with 1 mole of Cl2.
We found previously that 1.296 moles of CH4 reacted. Therefore, the number of moles of Cl2 reacted will also be 1.296 moles.
Using the molecular mass of Cl2 (70.90 g/mol), we can calculate the mass of Cl2:
mass of Cl2 = moles of Cl2 x molecular mass of Cl2
= 1.296 mol x 70.90 g/mol
= 91.91 g
Therefore, 91.91 grams of Cl2 reacted with CH4.
a) Convert 5.0g CH3Cl, 25.5 g CH2Cl2, and 589.0 g CHCl3 to mols, then convert EACH to mols CH4 used for that particular product. Total the g CH4 used, subtract from initial grams CH4, and convert that many g CH4 to mols CCl4 formed, then convert to grams CCl4.
b. Assume ALL of the CH4 is converted to CCl4. This is the theoretical yield.
c. %yield = (actual yield/theo yield)*100 = ?
d. You must mean Cl2 but you don't say which reaction. After you determine that it becomes a stoichiometry problem.