a,
DD A:
$Na+H_{2}O \to NaOH +1/2H_{2}$
$2Al+2NaOH +2H_{2}O \to 2NaAlO_{2}+3H_{2}$
$nH_{2}=\frac{6,72}{22,4}=0,3$
$CO_{2}+NaAlO_{2}+2H_{2}O \to NaHCO_{3}+Al(OH)_{3}$
$mAl(OH)_{3}=\frac{11,7}{78}=0,15$
Bảo toàn nguyên tố "Al"
$⇒nAl(OH)_{3}=nNaAlO_{2}=nAl=0,15$
Bảo toàn nguyên tố "Na"
$nNaAlO_{2}=nNaOH=nNa=0,15$
$mAl=0,15.27=4,05g$
$mNa=0,15.23=3,45g$
Trường hợp 1 : Nhôm tan hết trong ddNaOH
Chất rắn B:
$Fe+ 2HCl \to FeCl_{2}+H_{2}$
0,175 0,35 0,175
100ml=0,1lit
$nHCl=0,1.3,5=0,35$
$mFe=0,175.56=9,8g$
$mhh=mFe+mAl+mNa$
$⇔mhh=9,8+4,05+3,45=17,3g$
⇒Loại
Trường hợp 2: Al có dư trong chất rắn B:
$2Al+6HCl \to 2AlCl_{3}+3H_{2}$
$Fe+2HCl \to FeCl_{2}+H_{2}$
100ml=0,1lit
$mFe+mAl=14,45-4,05-3,45=6,95$
⇒$56nF+27nAl=6,95(1)$
$nHCl=0,1.3,5=0,35mol$
⇒$2nFe+3nAl=nHCl=0,35(2)$
(1)(2)⇒$\left \{ {{nFe=0,1} \atop {nAl=0,05}} \right.$
$mhh=mAl+mNa+mFe+mAl dư$
$⇔mhh=4,05+3,45+0,1.56+0,05.27=14,45g$(Nhận)
⇒$mAl=4,05+0,05.27=5,4g$
$mNa=3,45g$
$mFe=0,1.56=5,6g$
b,
$mddA=mAl+mH_{2}O+mNa-mH_{2}$
$⇔mddA=4,05+3,45+400-0,3.2=406,9g$
$C\%NaAlO_{2}=\frac{0,15.82}{406,9}.100=3,02\%$
$mddC=mddHCl+mAl dư+mFe-mH_{2}$
⇔$mddC=100.1,05+5,6+0,05.27-0,175.2=111,6g$
$C\%FeCl_{2}=\frac{0,1.127}{111,6}.100=11,38\%$
$C\%AlCl_{3}=\frac{0,05.133,5}{111,6}.100=5,98\%$
