How many grams of chloromethane will be obtained if 600 g of a mixture (N2 and CH4) enter into

How many grams of chloromethane will be obtained if 600 g of a mixture (N2 and CH4) enter into a reaction with Cl2. Mass fraction of CH4 = 20%.

Given:
m mixture (N2, CH4) = 600 g
ω (CH4) = 20%

To find:
m (CH3Cl) -?

Decision:
1) CH4 + Cl2 => CH3Cl + HCl;
2) m (CH4) = ω (CH4) * m mixture / 100% = 20% * 600/100% = 120 g;
3) M (CH4) = Mr (CH4) = Ar (C) * N (C) + Ar (H) * N (H) = 12 * 1 + 1 * 4 = 16 g / mol;
4) n (CH4) = m (CH4) / M (CH4) = 120/16 = 7.5 mol;
5) n (CH3Cl) = n (CH4) = 7.5 mol;
6) M (CH3Cl) = Mr (CH3Cl) = Ar (C) * N (C) + Ar (H) * N (H) + Ar (Cl) * N (Cl) = 12 * 1 + 1 * 3 + 35.5 * 1 = 50.5 g / mol;
7) m (CH3Cl) = n (CH3Cl) * M (CH3Cl) = 7.5 * 50.5 = 378.75 g.

Answer: The mass of CH3Cl is 378.75 g.