Bài 1:
1) Fe2(SO4)3 + 6BaCl2 ---> 2FeCl3 + 3BaSO4
2) FeCl3 + 3NaOH ---> Fe(OH)3 +3NaCl
3) 2Fe(OH)3 ---> Fe2O3 + 3H2O ($t^{o}$)
4) Fe2O3 + 6HCl ---> 2FeCl3 + 3H2O
Bài 2:
$n_{CaCO3}$ = 80÷100 = 0.8 (mol)
a) CaCO3 + HCl ---> CaCl2 + H2O + CO2
b)
Theo pt $n_{CO_{2} }$ = $n_{CaCO_{3} }$ = 0.8(mol)
$V_{CO2(đktc)}$ = 22.4 × 0.8 = 17.92 (L)
c)
Theo pt $n_{CaCl_{2} }$ = $n_{CaCO_{3} }$ = 0.08(mol)
=> m muối = 0.8 × 111 = 88.8(g)