Question

When 25.0 g of ch4 reacts completely with excess chlorine yielding 45.0 g of ch3cl, what is the percentage yield, according to ch4(g) + cl2(g) → ch3cl(g) + hcl(g)?

Answer 1

Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the reaction is,
    CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)

Since there is excess of Cl₂ gas, we can assume that all of CH₄ gas are reacted.

Moles of CH₄(g) = mass / molar mass
                           = 25.0 g / 16 g/mol
                           = 1.5625 mol

The stoichiometric ratio between CH₄(g) and CH₃Cl(g) is 1 : 1

Hence moles of CH₃Cl(g) = 1.5625 mol

Molar mass of CH₃Cl(g) = 50.5 g/mol
 
Mass of CH₃Cl(g) = number of moles x molar mass
                             = 1.5625 mol x 50.5 g/mol
                             = 78.9 g
Hence theoretical yield = 78.9 g
Actual yield = 45.0 g

Hence,
 Percentage yield = (45.0 g / 78.9 g) x 100% 
                             = 57.03%