Calculate the mass of aluminum that will be required for the complete separation of copper

Calculate the mass of aluminum that will be required for the complete separation of copper from 320 g of a 15% copper sulfate solution.

Given:

m (CuSO4) = 320 g

w% (CuSO4) = 15%

To find:

m (Al) -?

Solution:

2Al + 3CuSO4 = Al2 (SO4) 3 + 3Cu, – we solve the problem, relying on the composed reaction equation:

1) Find the mass of copper sulfate in the reacted solution:

m (CuSO4) = 320 g * 0.15 = 48 g

2) Find the amount of copper sulfate:

n (CuSO4) = m: M = 48 g: 160 g / mol = 0.3 mol

3) We compose a logical expression:

If 2 mol of Al requires 3 mol of CuSO4,

then x mol of Al will require 0.3 mol of CuSO4,

then x = 0.2 mol.

4) Find the mass of aluminum required for the reaction to proceed:

m (Al) = n * M = 0.2 mol * 27 g / mol = 54 g.

Answer: m (Al) = 54 g.