Question

2CH₄ + 4Cl₂ = 2CHCL₃ + 2HCL

1) If 5.5 CH₄ react with 27.4 g CL₂ what is the limiting reactant and what mass of HCl will form? (I can find the LR but I have problems with finding the mass of HCL)

2) If for the above reaction, the theoretical yield of CH₃Cl is 7.2g and the actual yield of HCl is 3.2 g, what is the percent yield?

Answer 1

   2CH₄ + 4Cl₂------->2CHCL₃ + 2HCL

no of moles of CH4 = W/G.M.Wt = 5.5/16 = 0.34375moles

no of moles of Cl2 = W/G.M.W     = 27.4/71   = 0.386 moles

2 moles of CH4 react with 4 molses of Cl2

0.34375 moles of CH4 react with = 4*0.34375/2   = 0.6875 moles of Cl2

Cl2 is limiting reagent

4 moles of Cl2 react with CH4 to form 2 moles of HCl

0.386 moles of Cl2 react with CH4 to form = 2*0.386/4 = 0.193 moles of HCl

mass of HCl = no of moles *gram molar mass

                      = 0.193*36.5   = 7.0445g

2. percent yield = Actual yield*100/theoritical yield

                          = 3.2*100/7.2   = 44.45%