8. Consider the following balanced reaction. Species Cu Molar Mass 159.61 g/mol

Question

8. Consider the following balanced reaction. Species Cu Molar Mass 159.61 g/mol 56.11 g/mol 97.56 g/mo CuSO4 (aq) + 2 KOH (aq) Cu(OH)2 (s) K2SO4 (aq) KOH Cu(OH a. How many moles of KOH is required to make 9.87 g of Cu(OH)? ,SO, 17426 gmol K SO4 Put a box around your answer. b. If 5.00 grams of CuSO4 is allowed to react with 6.00 grams of KOH what is the maximum amount of Cu(OH that can be fomed? Put a box around your answer Based on the information and your answer in part b, what would be the actual yield of this reaction if the % yield for the reaction was determined to be 65.5 %? c.

Explanation / Answer

CuSo4 + 2 KOH -----> Cu(OH)2 + K2SO4

a)

1 mole of Cu(OH)2 require 2 moles of KOH

97.56 g of Cu(OH)2 requires 2 * 56.11 g of KOH

9.87 g of Cu(OH)2 requires 2*56.11/(97.56)*9.87 g of KOH

                          = 11.3531 g of KOH

b)

1 mole of CuSO4 requires 2 moles of KOH

159.61 g of CuSO4 requires 2*56.11 g of KOH

5 g of CUSO4 requires 2*56.11/159.61*5 g of KOH

                     = 3.515 g of KOH

so CuSO4 is limiting reagent.

1 mole of CuSO4 produces 1 mole of Cu(OH)2

159.61 g of CuSO4 produces 97.56 g of Cu(OH)2

5 g of CuSO4 produces 97.56/159.61*5 g of Cu(OH)2

                     = 3.0562 g of Cu(OH)2

c)

Percentage yield = experimental yield/theritical yied*100

65.5 = x/3.0562*100

x = actual yield = 2.0018 g of Cu(OH)2

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