Fe+H2SO4->FeSO4+H2
0,2----0,2-------0,2-----0,2
m H2SO4=19,6g
->n H2SO4=\(\dfrac{19,6}{98}=0,2mol\)
=>mFe=0,2.56=11,2g
=>VH2=0,2.22,4=4,48l
FeSO4+2NaOH ->Fe(OH)2+Na2SO4
0,2--------0,4-----------0,2
CM NaOH=\(\dfrac{0,4}{0,3}=1,33M\)
m Fe(OH)2=0,2.90=18g