· For the reaction Cl2 + 2KBr → 2KCI + Brz, how many moles of potassium chloride are produced from 119 g of potassium bromide?

l

2

+

2

K

B

r

→

2

K

C

l

+

B

r

2

Given:

300

g

of

C

l

2

300

g

of

K

B

r

(a) Mass of

C

l

2

= 300g

Molecular mass of

C

l

2

= 2(35.45) = 70.906 g

/

m

o

l

No.of moles of

C

l

2

=

m

a

s

s

/

m

o

l

a

r

m

a

s

s

=

300

/

70.906

=

4.23

moles

Mass of

K

B

r

=

300

g

Molar mass of

K

B

r

=

39

+

79.9

=

118.9

g

/

m

o

l

No.of moles of

K

B

r

=

300

/

118.9

=

2.52

moles

Now,

Consider the equation:

C

l

2

+

2

K

B

r

→

2

K

C

l

+

B

r

2

1

mole of

C

l

2

= 2 moles of

K

C

l

4.23

moles of

C

l

2

= 2x4.23=

8.46

moles

2

moles of

K

B

r

= 2 moles of

K

C

l

2.52

moles of

K

B

r

=

2.52

moles

Now convert moles to mass,

Mass of

K

C

l

produced by

C

l

2

=

8.46

×

70.906

=

599.8

g

Mass of

K

C

l

produced by potassium bromide =

2.52

×

74.5

=

187.74

g

As

K

B

r

is producing lesser moles of product so,

K

B

r

is limiting reactant and produced potassium chloride=

187

Explanation: