64,0
Ta có:
Fe + 2HCl→ FeCl2 + H2
Fe3O4 + 8HCl → 2FeCl3 + FeCl2 + 4H2O
Bảo toàn Fe: nFe bđ = 0,2 + 0,2.3 = 0,8
2Fe → Fe2O3
0,8 → 0,4
=> m = 0,4.160 = 64 (g)
nếu sai mong bạn bỏ qua .
NaOH dư nên Al(OH)3 tan hết => chất rắn là Fe2O3
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\)
\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\)
\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)
\(FeCl_2+2NaOH\rightarrow e\left(OH\right)_2+2NaCl\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(4Fe\left(OH\right)_2+O_2+2H_2O\rightarrow4Fe\left(OH\right)_3\)
\(2Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\)
\(n_{FeCl2}=0,2+0,2=0,4\left(mol\right)\)
\(n_{FeCl3}=0,2.2=0,4\left(mol\right)\)
\(n_{Fe\left(OH\right)2}=n_{FeCl2}=0,4\left(mol\right)\)
\(n_{Fe\left(OH\right)3}=0,4+0,4=0,8\left(mol\right)\)
\(\Rightarrow n_{Fe2O3}=0,4\left(mol\right)\)
\(\Rightarrow m_{Fe2O3}=0,4.160=64\left(g\right)\)
