A student dissolved 0.946g Al in 35mL of 2.00M KOH and obtained 13.8g of dry alu

Question

A student dissolved 0.946g Al in 35mL of 2.00M KOH and obtained 13.8g of dry alum by following the experimental procedure given in this exercise. Assume that the Alum was 97.7% pure.

1. Calculate the theoretical and actual % yield.

2. Calculate the volume of excess KOH(aq) and the volume of 6.00M H2SO4(aq) needed to neutralize it.

3. Calculate the volume of 6.00M H2SO4(aq) needed just to precipitate Al(OH)3(s).

4. Calculate the volume of 6.00M H2SO4(aq) needed just to dissolve the Al(OH)3(s).

5. Calculate the total volume of 6.00M H2SO4(aq) required to carry out the entire process from start to finish.

Explanation / Answer

the chemical equation for the reaction is given by,

Al + KOH + 4 H2O -----> K[Al(OH)4] + 2 H2

2 K[Al(OH)4] + 4 H2SO4 + H2O------> K2SO4.Al2(SO4)3.24H2O

number of moles of Al = 0.946/27 = 0.03503 moles

number of moles of KOH = 35 * 10^-3 * 2 = 0.07 moles

So Al is limiting reagent.

1 mole of Al produces 1 mole of K[Al(OH)4] complex

0.03503 moles of Al produces 0.03503 moles of K[Al(OH)4]

and 2 moles of K[Al(OH)4] complex produces 1 mole of K2SO4.Al2(SO4)3.24H2O

0.3503 moles of K[Al(OH)4] complex produces 1/2*0.03503 moles of K2SO4.Al2(SO4)3.24H2O

mass of K2SO4.Al2(SO4)3.24H2O alum = 1/2*0.03503 * 948.77 = 16.6177g

Theoritical yield = 16.6177 g

since it is given alum is 13.8 g is 97.7% pure.

mass of Alum = 13.8*97.7/100 = 13.4826g

Percent of yield = actual yield/theoritical yield * 100 = 13.4826/16.6177*100 = 81.134%

Excess of KOH remaining = 0.07 - 0.03503 = 0.03497 moles

from the given 2 moles of KOH in 1 L of solution.

0.03497 moles of KOH present in 1/2 * 0.03497 L of solution.

                               = 0.01748 L = 17.48 mL

2 KOH + H2SO4 -------> K2SO4 + 2 H2O

2 moles of KOH requires 1 mole of H2SO4

0.03497 moles of KOH requires 1/2*0.03497 moles of H2SO4

                             = 0.01748 moles

from the given 6 moles of H2SO4 present in 1L of solution

0.01748 moles of H2So4 present in 1/6*0.01748 L of solution

                                = 0.002913 L = 2.913 mL

so 2.913 mL of H2SO4 is required to neutralize excess of KOH solution.

2 K[Al(OH)4] + 4 H2SO4 + 16 H2O------> K2SO4.Al2(SO4)3.24H2O

K[Al(OH)4] --------> KOH + Al(OH)3

so 2 moles of Al(OH)3 reacts 4 moles of H2SO4

0.03503 moles of Al(OH)3 reacts 4/2*0.03053 moles of H2SO4

                              = 0.06106 moles of H2SO4

6 moles of H2SO4 present in 1 L of solution

0.06106 moles of H2SO4 present in 1/6*0.06106 L of solution

                                 = 0.01018 L = 10.18 mL

so 10.18 mL of 6M H2SO4 is requires to react with Al(OH)3

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.