a)
\(n_{H_2SO_4}=0,5a\left(mol\right)\); nKOH = 2.0,2 = 0,4 (mol)
\(n_{Al\left(OH\right)_3}=\dfrac{0,39}{78}=0,005\left(mol\right)\)
TH1:
D chứa axit dư
PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_{\text{4}}+2H_2O\)
0,4------->0,2
=> \(n_{H_2SO_4\left(D\right)}=0,5a-0,2\left(mol\right)\)
PTHH: \(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)
0,005---->0,0075
=> 0,5a - 0,2 = 0,0075.2
=> a = 0,43
TH2: D chứa KOH dư
PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_{\text{4}}+2H_2O\)
a<-------0,5a
=> \(n_{KOH\left(D\right)}=0,4-a\left(mol\right)\)
PTHH: \(Al\left(OH\right)_3+KOH\rightarrow KAlO_2+2H_2O\)
0,005---->0,005
=> 0,4 - a = 0,005.2
=> a = 0,39
b)
Gọi số mol Fe3O4, FeCO3 là x, y (mol)
=> 232x + 116y = 2,688 (1)
TH1: a = 0,43
\(n_{H_2SO_4}=0,1.0,43=0,043\left(mol\right)\)
PTHH: \(Fe_3O_4+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+FeSO_4+4H_2O\)
\(FeCO_3+H_2SO_4\rightarrow FeSO_4+CO_2+H_2O\)
=> 4x + y = 0,043 (2)
(1)(2) => \(\left\{{}\begin{matrix}x=\dfrac{23}{2320}\left(mol\right)\\y=\dfrac{97}{29000}\left(mol\right)\end{matrix}\right.\) => \(\left\{{}\begin{matrix}m_{Fe_3O_4}=\dfrac{23}{2320}.232=2,3\left(g\right)\\m_{FeCO_3}=\dfrac{97}{29000}.116=0,388\left(g\right)\end{matrix}\right.\)
TH2: a = 0,39
=> \(n_{H_2SO_4}=0,39.0,1=0,039\left(mol\right)\)
=> 4x + y = 0,039 (3)
(1)(3) => \(\left\{{}\begin{matrix}x=\dfrac{459}{58000}\left(mol\right)\\y=\dfrac{213}{29000}\left(mol\right)\end{matrix}\right.\) => \(\left\{{}\begin{matrix}m_{Fe_3O_4}=\dfrac{459}{58000}.232=1,836\left(g\right)\\m_{FeCO_3}=\dfrac{213}{29000}.116=0,852\left(g\right)\end{matrix}\right.\)
