# Calculate the mass of the precipitate, which is formed by the interaction of 160 g of a 15%

**Calculate the mass of the precipitate, which is formed by the interaction of 160 g of a 15% solution of copper sulfate 2 with an excess of sodium hydroxide.**

Given:

m (CuSO4) = 160 g

w% (CuSO4) = 15%

To find:

m (draft) -?

Solution:

CuSO4 + 2NaOH = Cu (OH) 2 + Na2SO4, – we solve the problem based on the composed reaction equation:

1) Find the mass of copper sulfate that has reacted:

m (CuSO4) = 160 g * 0.15 = 24 g

2) Find the amount of copper sulfate:

n (CuSO4) = m: M = 24 g: 160 g / mol = 0.15 mol

3) We compose a logical expression:

If 1 mol of CuSO4 gives 1 mol of Cu (OH) 2,

then 0.15 mol of CuSO4 will give x mol of Cu (OH) 2,

then x = 0.15 mol.

4) Find the mass of precipitated copper hydroxide:

m (Cu (OH) 2) = n * M = 0.15 mol * 98 g / mol = 14.7 mol

Answer: m (Cu (OH) 2) = 14.7 mol.