Na+H2O→NaOH+1/2H2
a a 0,5a
Al+NaOH+H2O→NaAlO2+3/2H2
a a 1,5a
Rắn B: Al dư,Fe
Gọi nNa=a
nAl=b
nFe=c
nH2=0,448/22,4=0,02mol
→1,5a+0,5a=0,02
→2a=0,02
→a=0,01
mhh=23a+27b+56c=2,16
→27b+56c=1,93(1)
nAl dư=b-a
2Al+3CuSO4→Al2(SO4)3+3Cu
b-a 1,5(b-a)
Fe+CuSO4→FeSO4+Cu
c c
mCuSO4=200.4,8/100=9,6g
nCuSO4=9,6/160=0,06mol
nCu=3,2/64=0,05mol
→nCuSO4 dư 0,01mol
→1,5(b-a)+c=0,05
→1,5(b-0,01)+c=0,05
→1,5b-0,015+c=0,05
→1,5b+c=0,065(2)
Từ (1) và (2)
→b=0,03
c=0,02
mNa=0,01.23=0,23g
mAl=0,03.27=0,81g
mFe=0,02.56=1,12g
m rắn B=mAl dư+mFe=(0,03-0,01)27+0,02.56=1,66g
mdd C=200+1,66=201,66g
mAl2(SO4)3=0,01.342=3,42g
mFeSO4=0,02.152=3,04g
mCuSO4 dư=0,01.160=1,6g
C% Al2(SO4)3=3,42/201,66.100%=1,69%
C% FeSO4=3,04/201,66.100%=1,51%
C% CuSO4 dư=1,6/201,66.100%=0,79%
FeSO4+2NaOH→Fe(OH)2+Na2SO4
Al2(SO4)3+6NaOH→2Al(OH)3+3Na2SO4
CuSO4+2NaOH→Cu(OH)2+Na2SO4
4Fe(OH)2+O2→2Fe2O3+4H2O
2Al(OH)3→Al2O3+3H2O