\(1,m_{S}=m_{FeS}-m_{Fe}=17,6-11,2=6,4(g)\\ 2,m_{Na_2O}=m_{Na}+m_{O_2}=4,6+1,6=6,2(g)\\ 3,m_{H_2O}=m_{Ca(OH)_2}-m_{CaO}=37-28=9(g)\\ 4,m_{CaO}=m_{CaCO_3}-m_{CO_2}=100-44=56(g)\\ 5,\\ a,m_{Fe}+m_{H_2SO_4}=m_{FeSO_4}+m_{H_2}\\ b,m_{H_2}=m_{Fe}+m_{H_2SO_4}-m_{FeSO_4}=5,6+9,8-15,2=0,2(g)\)