Question

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?

Answer 1

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.

Answer 2

this is useless

Answer 3

Find tge product formed in these
1.CH4 + CL2=
2. C3H8 + O2