What mass of copper will be released during the interaction of 5.6 grams of iron 64 grams of copper II sulfate?

1. Let’s write down the equation of the proceeding reaction:

Fe + CuSO4 = Cu + FeSO4;

2.Calculate the chemical amount of the interacting iron:

n (Fe) = m (Fe): M (Fe) = 5.6: 56 = 0.1 mol;

3. find the amount of copper sulfate:

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

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

n (CuSO4) = 64: 160 = 0.4 mol;

4.In the absence of iron, we determine the amount of released copper:

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

5.Calculate the mass of copper:

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

Answer: 6.4 g.