What is the volume of hydrogen chloride formed during the complete chlorination of 112 l of methane?

bystudin | May 28, 2020 | Education

Explanation:

CH4 + 4Cl2 = CCl4 + 4HCl

From the conditions of the job, V (CH4) = 112 l. We calculate the amount of methane substance:

n (CH4) = V (CH4) / Vm = 112 / 22.4 = 5 mol.

In the reaction equation, methane has a coefficient of 1, and hydrogen chloride has a coefficient of 4. Based on this:

n (HCl) = n (CH4) ⋅ 4/1 = 5 ⋅ 4/1 = 20 mol;

V (HCl) = n (HCl) ⋅ Vm = 20 ⋅ 22.4 = 448 l.

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