Question

3. In the reaction between 80.5 g CuCl2 and 138 g KI, it produced CuI, KCl and I2. 70.4 g of I2 was collected from the reaction.


2 CuCl2 + 4 K= 2 CuI + 4 KCl + I2


a. Identify the limiting reactant.

b. Identify the excess reactant.

c. How much of the excess reactant was left over?

e. What is the maximum amount that can be made from the reaction?

d. What is the theoretical yield?
e. Compute for the percent yield.​

Answer 1

Answer:

CuSO

4

+4KI→2CuI+I

2

+2K

2

SO

4

2CuSO

4

+4KI⟶2CuI+I

2

+2K

2

SO

4

No. of Atom

No. of Atom

Cu

2

2

S

2

2

O

8

8

K

4

4

I

4

4

Answer 2

Answer:

a. KI

b. CuCl2

c. 24.6135 grams of CuCl2

d. 61.9 grams of KCl

e. 61.9 grams of KCl

Explanation:

a. Limiting Reagent

First calculate the number of moles given of the reactants.

• 80.5 g of CuCl2 is given. (n = given mass/molar mass)

n (Cucl2) = $\frac{80.5}{63.5 + 2(35.5)}$

n (CuCl2) = 0.5985

• 138g of KI is given.

n (KI) = $\frac{138}{39+127}$

n (KI) = 0.831

According to the balanced equation,

2 moles of CuCl2 require 4 moles of KI.

so, 1 mole of CuCl2 requires 2 moles of KI.

So, 0.5985 moles of CuCl2 requires approximately 1.2 moles of KI.

Which is not given to us, hence KI is the L.R.

b. Excess reactant

Since KI is the Limiting reactant, Cucl2 becomes the excess reactant.

c. Leftover Reactant

Since KI is the L.R. it will govern the reaction.

4 moles of KI require 2 moles of CuCl2

1 mole of KI requires 1/2 mole of CuCl2

0.831 moles of KI require 0.4155 moles of CuCl2.

We have 0.5985 moles of CuCl2.

Leftover moles of CuCl2 = 0.5985 - 0.4155

                                         = 0.183

Leftover grams of CuCl2 = leftover moles x molar mass

                                         = 0.183 x 134.5

                                         = 24.6135 g

Hence 24.6 grams of CuCl2 will be leftover.

d. Maximum amount of products

1 mole of KI gives 1/2 mole of CuI

1 mole of KI gives 1 mole of KCl

1 mole of KI gives 1/4 mole of I2.

Hence the maximum product formed will be KCl.

Moles of KCl formed = moles of KI used

                                  = 0.831

Mass of KCl formed = moles formed x molar mass

                                = 0.831 x (39 + 35.5)

                                = 0.831 x 74.5

                                = 61.9 grams

Hence KCl will be the maximum product to be made and it will be 61.9 grams.

e. Theoretical yield of KCl

1 mole KI gives 1 mole KCl

So, 0.831 moles of KI give 0.831 moles of KCl.

Theoretical Yield of KCl = 0.831 x 74.5

                                        = 61.9 grams

f. Percent yield of KCl

percentage yield = actual yield / theoretical yield x 100

Note= Actual yield comes from the experiment you perform in real.

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