[lelouch]

In a reaction between methane and chlorine, four products can be formed: CH3Cl, CH2Cl2, CHCl3, and CCl4. On one particular situation, 20.8 g of CH4 reacted, producing 5.0 g of CH3Cl, CH2Cl2 and 25.5 g 59.0 g CHCl3. All CH4 reacted. How many grams of CCl4 were formed? And how many grams of Cl2 reacted with CH4?

R: 63 g of CCl4, 269 g of Cl2

[Hunter2]

First you need for all reactions the equations.

Then you can calculate how many mol each product depends on. Need molecular weight each compound.

With the equation you can calculate the moles of chlorine and methane and backward the mass of it.

[Vidya]

Calculate the moles  of methane required from the gm of the compound formed over there
like CH3Cl
5.0 g CHCl3 X 1mole CHCl3 / molar mass CHCl3  X 1mole CH4/ 1mole CHCl3  X Molar mass CH4/ 1 mole of CH4
This will give you moles of CH4 was used for CHCl3 Similarly calculate moles of CH4 required for other compounds also
get the total number of moles of CH4 and subtract it from the total used up for different compounds
From left amount of moles of CH4 you can calculate moles CCl4 formed and finally grams also
Similarly set up calculations for the moles Cl2 required for each product and add all of them .Multiply total number of moles with molar mass to get the amount of Cl2 required .