I have problem with these 3 question in synthysis of aluminume : I would appreci

Question

I have problem with these 3 question in synthysis of aluminume : I would appreciate if you show calculation to find requested volumes:

Data:

mass of scrap Aluminume is 0.7423

Mass of Impurities is 0.2152

so Pure Al in scrap would be minus of these 2 = 0.5271 g

Mass of precipitate is 7.2042 g

Equitions:

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

K+ + Al3+ + 2 SO42- + 12 H2O ----> KAl(SO4).12H2O (S)

2AL(OH)3 + 3H2SO4 ---> 2AL3+ + 3SO4 + 6H2O

used 25 ml of 1.5 M KOH and 10 ml of 9M sulfuric acid

4. Calculate the mass of potassium aluminum sulfate produced. 5. Calculate the theoretical mass of potassium aluminum sulfate that could have been produced. (Hint: Remember to include the mass of the 12 waters of hydration when you are calculating the molar mass of your product). Calculate the percent yield of the potassium aluminum sulfate. 6. x 100% Percent Yield Actual ie Theoretical Yield 2. In this experiment volumes of 1.5 M potassium hydroxide and 9 M sulfuric acid were added, why was this information not precise?

Explanation / Answer

4) Here, the mass of KAl(SO4)212H2O produced = 7.2042 g (provided along with the problem)

5) Molar mass of KAl(SO4)212H2O = 474 grams/mol

Number of moles of Al in 0.5271 g = 0.5271 / 27 = 0.0195 mole

Mass of KAl(SO4)212H2O = 0.0195 mole x 474 g/mol = 9.243 g

Theoretical mass of KAl(SO4)212H2O that could have produced from 0.5271 g pure Al = 9.243 g

6) Percentage yield = (Actual yield/ Theoretical yield) x 100

= (7.2042/9.243) x 100 = 77.94 %

Part 2) Here, first Aluminum metal react with KOH to produce KAl(OH)4 with the evolution of hydrogen gas. Therefore, the number of moles of the KOH solution should greater than 0.0195 mole. This need not be precise and can be higher, because, the excess KOH will be neutralized with sulfuric acid. But if the number of moles of KOH is less than 0.0195, all the Aluminium will not react. Here, the number of moles of KOH is (25 x1.5)/1000= 0.0375.

When sulphuric acid is added to this solution, excess potassium hydroxide is neutralized. Also, the aluminate ions react with sulphuric acid to produce aluminum hydroxide. Further addition of the sulphuric acid dissolves the aluminum hydroxide. Therefore number of moles of sulfuric acid should be much higher, but need not be precise. Here, the number of moles of H2SO4 is (10 x 9)/1000= 0.09

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---