An excess of sodium hydroxide solution was added to 54 g of a solution with a mass fraction of iron (II) chloride of 20%.

An excess of sodium hydroxide solution was added to 54 g of a solution with a mass fraction of iron (II) chloride of 20%. Determine the mass of the precipitate.

1) Let’s compose the reaction equation FeCl2 + 2NaOH ==> Fe (OH) 2 | + 2NaCl
2) Determine the mass of ferric chloride in a solution of 54 g (solution) ==> 100% x g (FeCl2) ==> 20% therefore: m (FeCl2) = 20% * 54 g / 100% = 10.8 g
3) Determine the mass of the precipitate (Fe (OH) 2) 10.2 g (FeCl2) ==> 127 g / mol (FeCl2) x g (Fe (OH) 2) ==> 90 g / mol (Fe (OH) 2) therefore: m (Fe (OH) 2) = 90 g / mol * 10.2 g / 127 g / mol = 7.23 g
Answer: m (Fe (OH) 2) = 7.23 g