Giải thích các bước giải:
$a)\ n_{Fe}=\dfrac{5,6}{56}=0,1\ (mol)\\Fe+H_2SO_4\to FeSO_4+H_2\\n_{H_2}=n_{Fe}=0,1\ (mol)\\\to V=0,1×22,4=2,24\ (l)\\b)\ n_{H_2SO_4}=n_{Fe}=0,1\ (mol)\\m_{dd\ H_2SO_4}=\dfrac{0,1×98}{40\%}=24,5\ (g)\\m_{dd\ FeSO_4}=5,6+24,5-0,1×2=29,9\ (g)\\\to C\%_{dd\ FeSO_4}=\dfrac{0,1×152}{29,9}=50,8\%$