What volume of gas will be released when dissolving 1.28 g of copper in concentrated sulfuric acid?

Let’s write down the reaction scheme and arrange the coefficients:
Cu + 2H2SO4 = CuSO4 + SO2 + 2H2O
This is a redox reaction. Where Cu is a reducing agent and sulfur from sulfuric acid is an oxidizing agent. The odds are set using the electronic balance:
Cu0 – 2e = Cu + 2
S + 6 + 2e = S + 4
before SO2, the coefficient is 1, and before sulfuric acid – 2, since the second mole went to the binding of the copper ion.
Let us determine the amount of copper substance n (Cu) = m (Cu) / M (Cu), where M (Cu) = 64 g / mol is the molar mass of copper.
n (Cu) = 1.28 / 64 = 0.02 mol
According to the reaction equation, it can be seen that for 1 mol of copper there is 1 mol of SO2, that is, n (SO2) = n (Cu) = 0.02 mol.
Let us determine the volume of sulfur dioxide V (SO2) = n (SO2) * Vm, where Vm = 22.4 l / mol is the molar volume.
V (SO2) = 0.02 * 22.4 = 0.448 l