a,
$CuSO_4 + 2NaOH \rightarrow Cu(OH)_2+ Na_2SO_4$
$FeSO_4+ 2NaOH \rightarrow Fe(OH)_2+ Na_2SO_4$
$Fe_2(SO_4)_3+ 6NaOH \rightarrow 2Fe(OH)_3+ 3Na_2SO_4$
$Cu(OH)_2 \buildrel{{t^o}}\over\longrightarrow CuO+ H_2O$
$2Fe(OH)_2+ \frac{1}{2} O_2 \buildrel{{t^o}}\over\longrightarrow Fe_2O_3+ 2H_2O$
$2Fe(OH)_3 \buildrel{{t^o}}\over\longrightarrow Fe_2O_3+ 3H_2O$
$CuO+ CO \buildrel{{t^o}}\over\longrightarrow Cu+ CO_2$
$Fe_2O_3+ 3CO \buildrel{{t^o}}\over\longrightarrow 2Fe+ 3CO_2$
b,
mS= 60.(64/3)%= 12,8g
=> nS= n$SO_4^{2-}$= 12,8/32= 0,4 mol
=> m$SO_4^{2-}$= 0,4.96= 38,4g
=> m kim loại= 60-38,4= 21,6g= m