Calculate the mass of the precipitate that forms when 125 g of a 23% zinc chloride solution

Calculate the mass of the precipitate that forms when 125 g of a 23% zinc chloride solution is reacted with potassium hydroxide.

Given:

m (ZnCl2) = 125 g

w% (ZnCl2) = 23%

To find:

m (draft) -?

Decision:

ZnCl2 + 2KOH = 2KCl + Zn (OH) 2, – we will solve the problem, focusing on the composed reaction equation:

1) Find the mass of zinc chloride that has reacted:

m (ZnCl2) = 125 g * 0.23 = 28.75 g

2) Find the amount of zinc chloride:

n (ZnCl2) = m: M = 28.75 g: 136 g / mol = 0.21 mol

3) We compose a logical expression:

If 1 mol of ZnCl2 gives 1 mol of Zn (OH) 2,

then 0.21 mol of ZnCl2 will give x mol of Zn (OH) 2,

then x = 0.21 mol.

4) Find the mass of zinc hydroxide:

m (Zn (OH) 2) = n * M = 0.21 mol * 99 g / mol = 20.79 g.

Answer: m (Zn (OH) 2) = 20.79 g.