3. 1 mol Fe3O4.........3×56g Fe
2 moli Fe3O4.........x g Fe
x = 6×56 = 336g Fe
ms = 200g sol. HCl
C = 36,5%
C = md/ms×100 => md = C×ms/100 = 36,5×200/100 = 73g HCl
n = md/M = 73/36,5 = 2 moli HCl
Fe + 2HCl => FeCl2 + H2
56g Fe...........2 moli HCl
x = 56g Fe......2 moli HCl
2 moli HCl............22,4L H2
2 moli HCl.............x = 22,4L H2
