How much chlorine is required to react with 12 liters of methane?

Given:
V (CH4) = 12L
Find m (Cl2) -?
Decision:
Let’s compose the reaction equation:
CH4 + Cl2 = CH3Cl + HCl | CH4 – 1 mol Cl – 1 mol
1) Find v (nu – amount of substance) for CH4 using the formula:
v = V / Vm, Vm (molar volume) = 22.4 l / mol under normal conditions.
v (CH4) = 12 / 22.4 = 0.53 mol
2) v (CH4) = v (Cl2) = 0.53 mol Moles are equal due to the equal ratio in the reaction 1: 1
3) Find the molar mass for Cl2: M (Cl2) = 35.5 * 2 = 71 g / mol
4) Knowing M (Cl2) and v (Cl2), we find the mass m:
m = v * M
m (Cl2) = 71 * 0.53 = 37.63 g.
Answer: you need m (Cl2) = 37.63 g.