Question

In the reaction of 43.43 g of ethylene with 45.7 g of hydrogen chloride, 65.1 g of ethyl chloride was isolated by an experimenter. What was the percent yield? (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 tenths of a percent without units. H2C=CH2+ HCl → CH3CH2Cl

Answer 1

Molar mass of C2H4 = 28.0 g/mol

mass(C2H4)= 43.43 g

use:

number of mol of C2H4,

n = mass of C2H4/molar mass of C2H4

=(43.43 g)/(28.0 g/mol)

= 1.548 mol

Molar mass of HCl = 36.5 g/mol

mass(HCl)= 45.7 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(45.7 g)/(36.5 g/mol)

= 1.253 mol

Balanced chemical equation is:

C2H4 + HCl ---> C2H5Cl +

1 mol of C2H4 reacts with 1 mol of HCl

for 1.548 mol of C2H4, 1.548 mol of HCl is required

But we have 1.253 mol of HCl

so, HCl is limiting reagent

we will use HCl in further calculation

Molar mass of C2H5Cl= 64.5 g/mol

According to balanced equation

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

= (1/1)*1.253

= 1.253 mol

use:

mass of C2H5Cl = number of mol * molar mass

= 1.253*64.5

= 80.86 g

% yield = actual mass*100/theoretical mass

= 65.1*100/80.86

= 80.51%

Answer: 80.5