Question

What is the theoretical yield of ethyl chloride in the reaction of 37.10 g of ethylene with 58.33 g of hydrogen chloride? (For ethylene, MW=28.0 amu; for hydrogen chloride, MW=36.5 amu; for ethyl chloride, MW=64.5 amu.) Report your answer to the nearest hundredth of a gram without units. H2C=CH2 + HCl → CH3CH2Cl

Answer 1

Molar mass of C2H4, = 28.0 g/mol

mass(C2H4)= 37.1 g

use:

number of mol of C2H4,

n = mass of C2H4/molar mass of C2H4

=(37.1 g)/(28.0 g/mol)

= 1.323 mol

Molar mass of HCl = 36.5 g/mol

mass(HCl)= 58.33 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(58.33 g)/(36.5 g/mol)

= 1.6 mol

Balanced chemical equation is:

C2H4 + HCl ---> C2H5Cl +

1 mol of C2H4 reacts with 1 mol of HCl

for 1.323 mol of C2H4, 1.323 mol of HCl is required

But we have 1.6 mol of HCl

so, C2H4 is limiting reagent

we will use C2H4 in further calculation

Molar mass of C2H5Cl,

MM = 2*MM(C) + 5*MM(H) + 1*MM(Cl)

= 2*12.01 + 5*1.008 + 1*35.45

= 64.51 g/mol

According to balanced equation

mol of C2H5Cl formed = (1/1)* moles of C2H4

= (1/1)*1.323

= 1.323 mol

use:

mass of C2H5Cl = number of mol * molar mass

= 1.323*64.51

= 85.32 g

Answer: 85.32 g