16 g of copper was treated with 98% sulfuric acid solution m = 70 g, calculate the mass and volume

16 g of copper was treated with 98% sulfuric acid solution m = 70 g, calculate the mass and volume of the formed sulfur dioxide.

Given:
m (Cu) = 16 g
ω (H2SO4) = 98%
m solution (H2SO4) = 70 g

Find:
m (SO2) -?
V (SO2) -?

Solution:
1) Cu + 2H2SO4 => CuSO4 + SO2 ↑ + 2H2O;
2) M (Cu) = Mr (Cu) = Ar (Cu) = 64 g / mol;
M (H2SO4) = Mr (H2SO4) = Ar (H) * N (H) + Ar (S) * N (S) + Ar (O) * N (O) = 1 * 2 + 32 * 1 + 16 * 4 = 98 g / mol;
M (SO2) = Mr (SO2) = Ar (S) * N (S) + Ar (O) * N (O) = 32 * 1 + 16 * 2 = 64 g / mol;
3) n (Cu) = m (Cu) / M (Cu) = 16/64 = 0.25 mol;
4) m (H2SO4) = ω (H2SO4) * m solution (H2SO4) / 100% = 98% * 70/100% = 68.6 g;
5) n (H2SO4) = m (H2SO4) / M (H2SO4) = 68.6 / 98 = 0.7 mol;
6) n (SO2) = n (Cu) = 0.25 mol;
7) m (SO2) = n (SO2) * M (SO2) = 0.25 * 64 = 16 g;
8) V (SO2) = n (SO2) * Vm = 0.25 * 22.4 = 5.6 liters.

Answer: The mass of SO2 is 16 g; volume – 5.6 liters.