n H2SO4=78,4×20%÷98=0.16mol
2X + nH2SO4=> X2(SO4)n + nH2
m H2= (8,4+78,4)-86,5=0.3g
n H2=0.15 mol
n X=8,4÷X=0.15
=> X= 56(Fe)
n H2SO4 dư=0.16-0.15=0,01mol
H2SO4 + 2NaOH=> Na2SO4 + 2H2O
0.01______0.02
FeSO4 + 2NaOH=> Fe(OH)2 + Na2SO4
m NaOH còn lại =16-0.02×40=15.2g
n NaOH=0.38mol
n Fe(OH)2= 0.38÷2×90=17.1g