Al(OH)3+OH- == (AlO2)-+ H2O
n(AL3+)=△NaOH=10ml*10mol/L=0.1mol
n(AL2O3)=0.05mol
m(AL2O3)=0.05*102=5.1g
m(Fe2O3)=6-5.1=0.9g
Fe2O3+3H2SO4==Fe2(SO4)3+3H2O
160294
0.9g x1
x1=1.65375g
AL2O3+3H2SO4=AL2(SO4)3+3H2O
102 294
5.1x2
x2=14.7g
n(H2SO4)2=5/40 /2=0.0625mol
x1+x2=1.65375+14.7=16.35375g
n(H2SO4)=n1+n2=0.166875+0.0625=0.229375mol
c(H2SO4)=0.229375mol/0.1L=2.29375mol/L