Question

b) In the reaction of 277 g of carbon tetrachloride (CCl4) with an excess of hydrogen fluoride (HF), 187 g of CCl2F2 is obtained. What are the theoretical yield and percent yield for this reaction? CCl4Cg) 2 HF(g)CCl2F2(g) + 2 HCl(g)

Answer 1

Number of moles of CCl4 = 277 g / 153.82 g/mol = 1.80 mol

From the balanced equation we can say that

1 mole of CCl4 produces 1 mole of CCl2F2 so

1.80 mole of CCl4 will produce

= 1.80 mole of CCl4 *(1 mole of CCl2F2 / 1 mole of CCl4 )

= 1.80 mole of CCl2F2

mass of 1 mole of CCl2F2 = 120.91 g

so the mass of 1.80 mole of CCl2F2 = 218 g

Therefore, theoretical yield of CCl2F2 = 218 g

percent yield = (actual yield / theoretical yield )*100

percent yield = (187 / 218)*100

percent yield = 85.8

Therefore, percent yield of reaction = 85.8