Question

Each step in the process CH4+4Cl2⟶CCl4+4HCl CCl4+2HF⟶CCl2F2+2HCl has a yield of 60.0 %. The CCl4 formed in the first step is used as a reactant in the second step. If 6.50 mol of CH4 reacts, what is the total amount of HCl produced? Assume that Cl2 and HF are present in excess.

Answer 1

Reaction 1 : CH₄ + 4 Cl₂ CCl₄ + 4 HCl

Given : moles CH₄ = 6.50 mol

theoretical moles CCl₄ produced = moles CH₄ consumed

theoretical moles CCl₄ produced = 6.50 mol

actual moles CCl₄ produced = (theoretical moles CCl₄ produced) * (fractional yield)

actual moles CCl₄ produced = (6.50 mol) * (0.600)

actual moles CCl₄ produced = 3.90 mol

theoretical moles HCl produced = (moles CH₄ consumed) * (4 moles HCl / 1 mole CH₄)

theoretical moles HCl produced = (6.50 mol) * (4 / 1)

theoretical moles HCl produced = (6.50 mol) * 4

theoretical moles HCl produced = 26.0 mol

actual moles HCl produced = (theoretical moles HCl produced) * (fractional yield)

actual moles HCl produced = (26.0 mol) * (0.600)

actual moles HCl produced = 15.60 mol

Reaction 2 : CCl₄ + 2 HF CCl₂F₂ + 2 HCl

Given : moles CCl₄ = 3.90 mol

theoretical moles HCl produced = (moles CCl₄ consumed) * (2 moles HCl / 1 mole CCl₄)

theoretical moles HCl produced = (3.90 mol) * (2 / 1)

theoretical moles HCl produced = (3.90 mol) * 2

theoretical moles HCl produced = 7.80 mol

actual moles HCl produced = (theoretical moles HCl produced) * (fractional yield)

actual moles HCl produced = (7.80 mol) * (0.600)

actual moles HCl produced = 4.68 mol

Total moles HCl produced = (moles HCl produced in reaction 1) + (moles HCl produced in reaction 2)

Total moles HCl produced = (15.60 mol) + (4.68 mol)

Total moles HCl produced = 20.28 mol