$Fe+H_2SO_4\to FeSO_4+H_2\uparrow$
$n_{H_2}=\dfrac{11,2}{22,4}=0,5\ (mol)$
Theo phương trình:
$n_{Fe}=n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=0,5\ (mol)$
a, $m_{Fe}=0,5.56=28\ (g)$
b, $m_{H_2SO_4}=0,5.98=49\ (g)$
$m_{dd\ H_2SO_4}=\dfrac{49.100\%}{10\%}=490\ (g)$
c, $m_{muối}=m_{FeSO_4}=0,5.152=76\ (g)$
d, $m_{dd\ spứ}=m_{Fe}+m_{dd\ H_2SO_4}-m_{H_2}$
$=28+490-0,5.2$
$\to m_{dd\ spứ}=517\ (g)$
$C\%_{dd\ FeSO_4}=\dfrac{76}{517}.100\%=14,7\%$