\(m_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\
m_{H_2SO_4}=200.12,25\%=24,5g\\
n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\\
pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\\
LTL:0,2< 0,25=>H_2SO_4\left(d\text{ư}\right)Fe\left(h\text{ết}\right)\)
\(n_{H_2}=n_{Fe}=0,2\left(mol\right)\\
V_{H_2}=0,2.22,4=4,48l\)
\(C\%=\dfrac{11,2}{24,5}.100\%=45,7\%\)