Pre-Lab Section Case Solution And Analysis, HBR Case Study Solution & Analysis of Harvard Case Studies

Pre-Lab Section Case Solution

Prelab Questions and Answers

A dance hall contains 173 men and 238 women. A certain “promenade” calls for the formation of groups, with each group consisting of one man with two women.

How many groups could be formed? (Show calculation)

Ans:

No. of groups men can form:

173 / 1 = 173

No. of groups women can form:

238 / 2 = 119

119 < 173, so it is clear that we can form 119 groups has one man and two women.

Which sex is in excess? By how much? (Show calculations)

Ans:

From the above, we can see that men are in excess.

173 - 119 = 54

There are 54 extra men.

The Foul Air Power and Light Company has been ordered to remove sulfur dioxide from its smoking stacks. Management decides to pass the gas ovet solid calcium oxide thereby forming solid calcium sulfide.

CaO + SO₂à CaSO₃

In a 24-hr period, 2500 moles of SO₂ pass over 85,000 moles of CaO. Show calculations for the following:

What is the limiting reagent?

Ans:

1 mole CaSO₃

2,500 moles S O₂ x = 2,500 moles CaSO₃

1 mole SO₂

1 mole CaSO₃

85,000 moles CaO x = 85,000 moles CaSO₃

1 mole CaO

The limiting reagent is S O₂.

Which reagent is in excess?

Ans: CaO.

How much excess reagent remains?

Ans:

1 mole CaO

2,500 moles S O₂ x = 2,500 moles CO (really reacted)

1 mole SO₂

85,000 moles CaO Ca: 1x40=40

2,500 moles reacted O : 1x16=16

= 82,500 moles in excess = 56g Cao = 1moles CaO

56g gCaO

82,500 moles in excess x = 462000g = 4.62x10⁵gCaO

1 moles CaO

How much product can be formed?

Ans:

Ca: 1x40 = 40

O: 3x16 = 48

S: 1x32 = 32

= 120g CaSO₃ = 1 moles CaSO₃

120 gCaSO₃

2,500 moles CaS O₃ x = 300,000g CaSO₃

1 mole CaSO₃

Experimental Section

Procedure

Obtain a clean 50 mL beaker, a 150 mL beaker, a funnel and a glass stirring rod.

Obtain between 0.4g and 0.8g of aluminum sulfate (Al₂(SO₄)₃) using a weighing cup and a scale.

Transfer the substance completely into the 150 mL beaker using distilled water if needed.

Add about 50 mL of distilled water and 10 mL of nitric acid to the beaker of aluminum sulfate and stir with a stirring rod.

Use a graduated cylinder to transfer exactly 10.0 mL of aqueous barium chloride solution to the 50 mL beaker and make sure to take note of the concentration listed on the label.

Using the hot plate, heat the aluminum sulfate solution until condensation appears on the sides of the beaker but be very careful to not let the solution boil.

Using a clean eye dropper, add the barium chloride solution slowly and steadily to the aluminum sulfate solution. During the process, you should be continuously stirring. If you are doing this properly, you should notice that the solution is turning cloudy.

NOTE: This must be done SLOWLY, and the solution must be kept hot throughout this process.

When you can no longer extract the barium chloride solution with the eye-dropper, rinse the beaker and eye- dropper with small amounts of distilled water and add this to the aluminum sulfate solution.

Keep the mixture warm for the next ten minutes, being very careful NOT TO STIR and NOT TO ALLOW A BOIL.

As the mixture is cooling, begin preparing the filtration described as follows:

Acquire a circular filter and record its mass.

Fold it twice, halving it each time. This should result in a cone-looking shape that you will position in a clean funnel over a beaker to help filter your substance.

Once the contents of your mixture have appropriately cooled, pour the mixture over the filter and allow all of the liquid to drain.

Once all of the liquid has drained, carefully remove the filter and store it according to your instructor's guidelines. The following week, record the mass of the paper and the precipitate for the following calculations..................

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Pre-Lab Section Harvard Case Solution & Analysis