A solution containing 240 g of sodium hydroxide was added to a solution containing

A solution containing 240 g of sodium hydroxide was added to a solution containing an excess of iron (III) chloride. Determine the mass and amount of the formed iron (III) hydroxide.

Given:
m (NaOH) = 240 g

To find:
n (sediment) -?
m (draft) -?

Decision:
1) FeCl3 + 3NaOH => 3NaCl + Fe (OH) 3 ↓;
2) M (NaOH) = Mr (NaOH) = Ar (Na) + Ar (O) + Ar (H) = 23 + 16 + 1 = 40 g / mol;
M (Fe (OH) 3) = Mr (Fe (OH) 3) = Ar (Fe) + Ar (O) * 3 + Ar (H) * 3 = 56 + 16 * 3 + 1 * 3 = 107 g / mole;
3) n (NaOH) = m (NaOH) / M (NaOH) = 240/40 = 6 mol;
4) n (Fe (OH) 3) = n (NaOH) / 3 = 6/3 = 2 mol;
5) m (Fe (OH) 3) = n (Fe (OH) 3) * M (Fe (OH) 3) = 2 * 107 = 214 g.

Answer: The amount of the substance Fe (OH) 3 is 2 mol; weight – 214 g.