alum [KAL(SO4)2 times xH2O] is used in food preparation, dye fixation and water

Question

alum [KAL(SO4)2 times xH2O] is used in food preparation, dye fixation and water purification. To prepare alum, aluminum is reacted with potassium hydroxide and the product with sulfuric acid.

1) A 0.6815g sample of alum is heated to drive off the waters of hydration, and the resulting KAL(SO4)2 weighs 0.3709 g. Determine the value of x AND the complete formula of alum.

2) when 0.7939g of aluminum is used, 8.570g of alum forms. what is the percent yield?

Explanation / Answer

For 2), we make several equations Al + 2KOH + 2H2O --> 2K(+) + [Al(OH)4](-) + H2 2H(+) + 2[Al(OH)4](-) --> 2Al(OH)3 + 2H2O Al(OH)3 + 3H(+) + 3H2O --> [Al(H2O)6](3+) K(+) + [Al(H2O)6](3+) + 2SO4(2-) + 6H2O --> KAl(SO4)2 * 12H2O The (+) or (-) is the charge. So: the mole ratio between starting Aluminum and ending alum is : 1 mol Al :1 mol [Al(OH)4](-) 2 mol [Al(OH)4](-) :2 mol Al(OH)3 1 mol Al(OH)3 : 1 mol [Al(H2O)6](3+) 1 mol [Al(H2O)6](3+) : 1 mol KAl(SO4)2 * 12H2O Which simplifies to 1 mol Al : 1 mol alum Al=26.9815386 g/mol so, (0.7939 g)*(mol/26.9815386 g)=0.0294238 mol Al So 0.0294238 mol alum is EXPECTED to form which is: (0.0.0294238 mol)*(258.2 g/mol)=7.59723g alum EXPECTED 8.57 g alum actually forms so % yeild is: (actual yeild) / (theoretical or expected yeild) = = 8.570 g / 7.59723 g =1.12804 = 112.804 %

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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