Question

32.00 mL of 0.1900 M of Al (OH)3 react with 30.00 mL of 0.2500 M H2SO4 . Calculate the mass of the precipitate.

Answer 1

Solution:

The reaction of Al(OH)3 with H2SO4 to form Al2(SO4)3 precipitate can be given as,

2Al(OH)3 + 3H2SO4 ===> Al2(SO4)3 + 6H2O

>> Number of moles of Al(OH)3 = Molarity x Volume in L

= 0.1900 M x 0.032 L = 0.00608 mol

>> Number of moles of H2SO4 = Molarity x Volume in L

= 0.2500 M x 0.030 L = 0.00750 mol

Since, 2 mol of Al(OH)3 requires 3 mol of H2SO4.

Hence, 0.00608 mol will required = 0.00608 mol x 3 /2

= 0.00912 mol of H2SO4

But the actual number of mol of H2SO4 (0.00750 mol) is less than required.

>> Hence, H2SO4 will be the limiting reagent.

From equation, it can be seen that 3 mol H2SO4 forms 1 mol Al2(SO4)3

Hence, 0.00750 mol of H2SO4 will form = 0.00750 mol x 1/3

= 0.00250 mol of Al2(SO4)3

Thus,

Mass of Al2(SO4)3 = Number of mol x molar mass

= 0.00250 mol x 342.15 g mol-1 = 0.855 g