Suppose that each of these reactions has a 90% yield. 

CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)

CH₃Cl(g) + Cl₂(g) → CH₂Cl₂(l) + HCl(g)

Starting with 50.0 g CH₄ in the first reaction and an excess of Cl₂(g), the number of grams of CH₂Cl₂ formed in the second reaction is

(a) 50.0 x 0.81 x (85/16)

(b) 50.0 x 0.90

(c) 50.0 x 0.90 x 0.90

(d) 50.0 x 0.90 x 0.90 x (16/50.5)(70.9/85)