8. How many grams of aluminum hyfroxide will react with 75.5 ml of 0.250 M sulfu
ID: 906908 • Letter: 8
Question
8. How many grams of aluminum hyfroxide will react with 75.5 ml of 0.250 M sulfuric acid?
Al(OH)3 + H2SO4 > Al2(SO4)3 + H20 (unbalanced)
a. Detrmine the number of mole (mmoles) sulfuric acid used.
b. Determine the number of moles (mmoles) Aluminum hydroxide reacted.
c. Detrmine the mass (grams) of aluminum hydroxide reacted.
9. Consider the following solutions. Put in order according to the boiling points from highest point to lowest. (Hint: Colligative properties. Kb=0.52 C/m for water)
a. 0.1 M c6h12o6
b. 0.1 M CaCl2
c. 0.1 M NaCl
Explanation / Answer
(8)
2Al(OH)3 + 3H2SO4 ------------> Al2(SO4)3 + 6H2O
No .of moles of H2SO4 = conc. x volume = 0.250 x 0.0755 = 0.019 moles.
According to the balance equation,
3 moles of H2SO4 reacts with 2 moles ( 2 x 78 = 156 g) of Al(OH)3
=> No.of grams of Al(OH)3 that will react with 0.019 moles ( 1.86 g) of H2SO4 = 0.019 x 156 / 3 = 0.99 g
(9)
deltaTb = i x m x Kb
=> delta Tb directly proportional to i ( vantHoff factor)
Therefore, boiling point order is,
0.1 M CaCl2 > 0.1 M NaCl > 0.1 M c6h12o6