What mass of copper will be released when 5.4 g of aluminum is exposed to a solution

What mass of copper will be released when 5.4 g of aluminum is exposed to a solution containing 16 g of copper sulfate?

1. Aluminum, being a more active metal, displaces copper from its sulfate solution:

2Al + 3CuSO4 = 3Cu + Al2 (SO4) 3;

2. Let’s calculate the chemical amounts of aluminum and copper sulfate:

n (Al) = m (Al): M (Al) = 5.4: 27 = 0.2 mol;

n (CuSO4) = m (CuSO4): M (CuSO4);

M (CuSO4) = 64 + 32 + 4 * 16 = 160 g / mol;

n (CuSO4) = 16: 160 = 0.1 mol;

3. Copper sulfate is taken in the deficiency, we determine from its data the amount of copper obtained:

n (Cu) = n (CuSO4) = 0.1 mol;

4. Find the mass of copper:

m (Cu) = n (Cu) * M (Cu) = 0.1 * 64 = 6.4 g.

Answer: 6.4 g.