What volume of gas will be released during the interaction of 1.08 g of silver and 196 g of 70%

What volume of gas will be released during the interaction of 1.08 g of silver and 196 g of 70% concentrated sulfuric acid solution?

Given:
m (Ag) = 1.08 g
m solution (H2SO4) = 196 g
ω (H2SO4) = 70%

Find:
V (gas) -?

Solution:
1) 2Ag + 2H2SO4 (conc.) = (TOC) => Ag2SO4 + SO2 ↑ + 2H2O;
2) M (Ag) = Mr (Ag) = Ar (Ag) = 108 g / mol;
M (H2SO4) = Mr (H2SO4) = Ar (H) * 2 + Ar (S) + Ar (O) * 4 = 1 * 2 + 32 + 16 * 4 = 98 g / mol;
3) m (H2SO4) = ω (H2SO4) * m solution (H2SO4) / 100% = 70% * 196/100% = 137.2 g;
4) n (H2SO4) = m (H2SO4) / M (H2SO4) = 137.2 / 98 = 1.4 mol;
5) n (Ag) = m (Ag) / M (Ag) = 1.08 / 108 = 0.01 mol;
6) n (SO2) = n (Ag) / 2 = 0.01 / 2 = 0.005 mol;
7) V (SO2) = n (SO2) * Vm = 0.005 * 22.4 = 0.112 liters.

Answer: The volume of SO2 is 0.112 liters.