Question

Part A How many grams of dichloromethane result from reaction of 1.71 kg of methane if the yield is 45.6 %? CH, (g) + 2Cl2 (g) - CH2Cl (1 + 2HC13) Methane Chlorine Dichloroch IVO AZP on o ? Submit Resuest Answer

Answer 1

The reaction is

CH4 +2Cl2 ----------> CH2Cl2 + 2HCl

From this

1 mole ( 16g/mol) of methane produce 1 mole ( 85g/mol) of CH2Cl2

If 1.71kg = 1.71 x1000 g of methane is used , the mass of CH2Cl2 = 1.71x1000g x 85g/mol / 16g/mol

=9084.38 g

thus the theoretical yield of CH2Cl2 = 9084.38g

However the yield was only 45.6%

Thus the actual yield of CH2Cl2 = theoretical yield x % yield

= 9084.38g x 45.6/100 =4142.47 g