When 20.5 g of methane and 45.0 g of chlorine gas undergo a reaction that has a 75.0% yield, what mass (g) of chloromethane (CH3Cl) forms? Hydrogen chloride also forms.

Answer 1

CH4 + Cl2 ==> CH3Cl + HCl
mols CH4 = g/molar mass = 20.5/16 = about 1.3 (you need to clean up my estimates).
mols Cl2 = 45/71 = about 0.64g

mols CH3Cl formed from CH4 = 1.28
mols CH3Cl formed from Cl2 = 0.64
Obviously both numbers can't be right; the correct value in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent. Thus Cl2 is the limiting reagent and 0.64 mols CH3Cl will be formed.
If you have a 75% yield, the amount ACTUALLY formed will be 0.64 x 0.75 = ?g