Sodium hydroxide solution was added to a 4% solution of aluminum sulfate weighing 171 g until

Sodium hydroxide solution was added to a 4% solution of aluminum sulfate weighing 171 g until a precipitate was formed. If we assume that the sulfate has completely reacted, then what mass of aluminum hydroxide was formed?

Given:
ω (Al2 (SO4) 3) = 4%
m solution (Al2 (SO4) 3) = 171 g

To find:
m (Al (OH) 3) -?

Decision:
1) Al2 (SO4) 3 + 6NaOH => 2Al (OH) 3 ↓ + 3Na2SO4;
2) m (Al2 (SO4) 3) = ω * m solution / 100% = 4% * 171/100% = 6.84 g;
3) n (Al2 (SO4) 3) = m / M = 6.84 / 342 = 0.02 mol;
4) n (Al (OH) 3) = n (Al2 (SO4) 3) * 2 = 0.02 * 2 = 0.04 mol;
5) m (Al (OH) 3) = n * M = 0.04 * 78 = 3.12 g.

Answer: The mass of Al (OH) 3 is 3.12 g.