What mass of a 75% sulfuric acid solution is required to interact with 3.2 g copper?

Due to the fact that copper is to the right of hydrogen in the series of metal voltages, the interaction will not work with an ordinary solution of sulfate acid, but only with a concentrated one. Then the reaction will look like this:

Cu + 2H2SO4 (conc) = CuSO4 + SO2 + 2H2O.

M (Cu) = 63.546 g / mol.

M (H2SO4) = 98 g / mol.

Find the amount of copper:

n (Cu) = m / M = 3.2 / 63.546 = 0.05 mol.

According to the stoichiometric coefficients, the amount of sulfuric acid in the reaction is twice as much as copper.

n (H2SO4) = 2n (Cu),

n (H2SO4) = 2 * 0.05 = 0.01 mol.

m (H2SO4) = M * m = 0.01 * 98 = 9.8 g.

It is pure sulfuric acid in solution. It is 75% of the desired solution.

m (solution) = m (H2SO4) / w = 9.8 / 0.75 = 13.066 g.