Answer: m Fe(OH)3 = 0.165 mol FeCl3 x (1 mol Fe(OH)3 /1 mol FeCl3 ) x (106.88 g Fe(OH)3 / 1 mol Fe(OH)3 ) = 17.7 g Fe(OH)3 m NaCl = 0.165 mol FeCl3 x (3 mol NaCl / 1 mol FeCl3 ) x (58.44 g NaCl / 1 mol NaCl ) = 29.0 NaCl Therefore sodium hydroxide is in excess by 1.7 g, the mass of iron (III) hydroxide produced is 17.7 g and the mass of sodium chloride produced is 29.0g.
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