Calculate the mass of the precipitate, which is formed by the interaction of ferrum (III) sulfate, the mass of which

Calculate the mass of the precipitate, which is formed by the interaction of ferrum (III) sulfate, the mass of which is 1.6 g, with a solution of barium hydroxide, the mass of which is 20 g, if the mass fraction of alkali is 15%.

Given:

m (Fe2 (SO4) 3) = 1.6 g

m (Ba (OH) 2) = 20 g

w% (Ba (OH) 2) = 15%

Find:

m (draft) -?

Solution:

3Ba (OH) 2 + Fe2 (SO4) 3 = 3BaSO4 + 2Fe (OH) 3, – we solve the problem based on the composed reaction equation:

1) Find the mass of barium hydroxide in solution:

m (Ba (OH) 2) = 20 g * 0.15 = 3 g

2) Find the amount of ferrous sulfate (3) and barium hydroxide:

n (Fe2 (SO4) 3) = m: M = 1.6 g: 400 g / mol = 0.004 mol

n (Ba (OH) 2) = m: M = 3 g: 171 g / mol = 0.0175 mol

We start from a lower value to get more accurate calculations. We work with Fe2 (SO4) 3:

3) We compose a logical expression:

if 1 mol of Fe2 (SO4) 3 gives 3 mol of BaSO4,

then 0.004 mol of Fe2 (SO4) 3 will give x mol of BaSO4,

then x = 0.012 mol.

4) Find the mass of barium sulfate precipitated during the reaction:

m (BaSO4) = n * M = 0.012 mol * 171 g / mol = 2.796 g.

Answer: m (BaSO4) = 2.796 g.