Consider the reactionCH4(g) + CCl4(g) → 2CH2Cl2(g) Given an initial mass of 17.02 g CH4, an excess of CCl4, and assuming that all of the reactant is converted to product(s), and none is lost, calculate the mass (g) of CH2Cl2 produced by the reaction.?

Molar mass of CH₄ = (12.01 + 1.008×4) g/mol = 16.04 g/mol
Molar mass of CH₂Cl₂ = (12.01 + 1.008×2 + 35.45×2) g/mol = 84.93 g/mol

Assume that the reaction occurs:
CH₄(g) + CCl₄(g) → 2CH₂Cl₂
Mole ratio CH₄ : CH₂Cl₂ = 1 : 2

Method 1:
No. of moles of CH₄ reacted = (17.02 g) / (16.04 g/mol) = 1.061 mol
No. of moles of CH₂Cl₂ produced = (1.061 mol) × (2/1) = 2.122 mol
Mass of CH₂Cl₂ produced = (2.122 mol) × (84.93 g/mol) = 180.2 g

Method 2:
(17.02 g CH₄) × (1 mol CH₄ / 16.04 CH₄) × (2 mol CH₂Cl₂ / 1 mol CH₄) × (84.93 g CH₂Cl₂ / 1 mol CH₂Cl₂)
= 180.2 g CH₂Cl₂

No such reaction