1 phần \(\left\{\begin{matrix} K: a \\ Al:b \\ Fe: c \end{matrix}\right.\)

(II) H₂O dư

\(\\ K\rightarrow KOH + \frac{1}{2}H_2 \\ a \hspace{30pt}a \hspace{40pt}\frac{a }{2}\)

\(\\Al + OH^-+H_2O\rightarrow AlO_2+\frac{3}{2}H_2 \\ b>a \hspace{130pt} \frac{3a}{2}\)

(I) Ba(OH)₂ dư (H₂O dư)

\(\\ K \rightarrow KOH+\frac{1}{2}H_2 \\ a \hspace{30pt}a \hspace{40pt}\frac{a}{2}\)

\(\\Al + OH^-+H_2O\rightarrow AlO_2+\frac{3}{2}H_2 \\ b\hspace{25pt}b \hspace{120pt} \frac{3b}{2}\)

\(nH_2(I)=\frac{a}{2}+\frac{3b}{2}=0,275 \ mol\)

\(nH_2(II)=0,2 \ mol <0,275\)

\(\Rightarrow\) phần II. Al dư

\(0,2=\frac{a+3a}{2}\Rightarrow a=0,1\)

\(\Rightarrow b= 0,15\)

chất không tan \(\left\{\begin{matrix} Al_{du} \\ Fe: C \end{matrix}\right.\overset{HCl}{\rightarrow} \left\{\begin{matrix} \frac{3}{2}H_2 \\ H_2 \end{matrix}\right.\)

Al dư= 0,15 - 0,1 = 0,05

\(nH_2(HCl)=0,175=0,05.\frac{3}{2}+C\)

\(\Rightarrow C=0,1\)

K: 3,9 (g)

Al: 4,05 (g)

Fe: 5,6 (g)