Respuesta :

Lookslikecherry05

Answer: 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:

CintiaSalazar

Taking into account the reaction stoichiometry, 1 mol of KCl is formed from 119 grams of KBr.

Reaction stoichiometry

In first place, the balanced reaction is:

Cl₂ + 2 KBr → 2 KCI + Br₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Cl₂: 1 mole
  • KBr: 2 moles
  • KCI: 2 moles
  • Br₂: 1 mole

The molar mass of the compounds is:

  • Cl₂: 70.9 g/mole
  • KBr: 119 g/mole
  • KCI: 74.55 g/mole
  • Br₂: 159.8 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Cl₂: 1 mole ×70.9 g/mole= 70.9 grams
  • KBr: 2 moles ×119 g/mole= 238 grams
  • KCI: 2 moles ×74.55 g/mole= 149.1 grams
  • Br₂: 1 mole ×159.8 g/mole= 159.8 grams

Moles of potassium chloride produced

The following rule of three can be applied: if by reaction stoichiometry 238 grams of KBr form 2 moles of KCl, 119 grams of KBr form how many moles of KCl?

[tex]moles of KCl=\frac{119 grams of KBrx2 moles of KCl}{238 grams of KBr}[/tex]

moles of KCl= 1 mole

Then, 1 mol of KCl is formed from 119 grams of KBr.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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