Calculate the mass of iron required for the reaction with 980 g of a 15% sulfuric acid solution.

Calculate the mass of iron required for the reaction with 980 g of a 15% sulfuric acid solution. How much hydrogen will be released during this?

Given:
m solution (H2SO4) = 980 g
ω (H2SO4) = 15%

To find:
m (Fe) -?
V (H2) -?

Decision:
1) Fe + H2SO4 => FeSO4 + H2;
2) Mr (Fe) = Ar (Fe) = 56 g / mol;
Mr (H2SO4) = Ar (H) * 2 + Ar (S) * 4 + Ar (O) * 4 = 1 * 2 + 32 + 16 * 4 = 98 g / mol;
3) m (H2SO4) = ω (H2SO4) * m solution (H2SO4) / 100% = 15% * 980/100% = 147 g;
4) n (H2SO4) = m (H2SO4) / Mr (H2SO4) = 147/98 = 1.5 mol;
5) n (Fe) = n (H2SO4) = 1.5 mol;
6) m (Fe) = n (Fe) * Mr (Fe) = 1.5 * 56 = 84 g;
7) n (H2) = n (H2SO4) = 1.5 mol;
8) V (H2) = n (H2) * Vm = 1.5 * 22.4 = 33.6 liters.

Answer: The mass of Fe is 84 g; volume H2 – 33.6 liters.