React 26.7 g of aluminum chloride and 12 g of sodium hydroxide. The mass of the sediment obtained?

Given:
m (AlCl3) = 26.7 g
m (NaOH) = 12 g

Find:
m (draft) -?

Solution:
1) n (AlCl3) = m (AlCl3) / M (AlCl3) = 26.7 / 133.5 = 0.2 mol;
2) n (NaOH) = m (NaOH) / M (NaOH) = 12/40 = 0.3 mol;
3) AlCl3 + 3NaOH => Al (OH) 3 ↓ + 3NaCl;
4) n (Al (OH) 3) = n (NaOH) / 3 = 0.3 / 3 = 0.1 mol;
5) m (Al (OH) 3) = n (Al (OH) 3) * M (Al (OH) 3) = 0.1 * 78 = 7.8 g.

Answer: The mass of Al (OH) 3 is 7.8 g.