(k biết cách nào đúng, chắc C1 đấy)
C1:
\(n_{HCl}=0,4.1,5=0,6\left(mol\right)\\ PTHH:Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ m_{Fe_2O_3}=0,1.160=16\left(g\right)\\ V_{ddspu}=0,4\left(l\right)\\ C_{M_A}=\frac{0,2}{0,4}=0,5\left(M\right)\)
C2:
\(n_{HCl}=0,4.1,5=0,6\left(mol\right)\\ PTHH:Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ m_{Fe_2O_3}=0,1.160=16\left(g\right)\\ V_{ddspu}=0,1.22,4+0,4=2,64\left(l\right)\\ C_{M_A}=\frac{0,2}{2,64}=0,08\left(M\right)\)
nHCl = 0.6 mol
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
0.1______0.6______0.2
mFe2O3 = 0.1*160 = 16 g
CM FeCl3 = 0.2/0.4 = 0.5M