1)

\(n_{Al}^{} = \frac{{5,94}}{{27}} = 0,22\;{\text{mol}}\)

\(Al + H_2^{}O + NaOH \to NaAlO_2^{} + \frac{3}{2}H_2^{}\)

\(n_{H_2^{}}^{} = 0,22.\frac{3}{2} = 0,33\;{\text{mol}}\)

\(n_{KMnO_4^{}}^{} = \frac{{1,896}}{{158}} = 0,012\)

\(2KMnO_4^{} + 16HCl \to 2KCl + 2MnCl_2^{} + 5Cl_2^{} + 8H_2^{}O\)

\({\text{  0}}{\text{,012                                                    }}\frac{{0,012.5}}{2} = 0,03\)

\(n_{Cl_2^{}}^{} = 0,03{\text{ mol}}\)

\(n_{KClO_3^{}}^{} = \frac{{12,25}}{{122,5}} = 0,1\;{\text{mol}}\)

\(KClO_3^{}\xrightarrow{{{t^o}}}KCl + \frac{3}{2}O_2^{}\)

\(\;0,1{\text{                           0}}{\text{,15}}\)

\({\text{Dot khi H}}_2^{},Cl_2^{},O_2^{}{\text{ trong cung 1 dung dich}}\)

\(2H_2^{} + O_2^{}\xrightarrow{{{t^o}}}2H_2^{}O\)

\(H_2^{} + Cl_2^{}\xrightarrow{{{t^o}}}2HCl\)

\(n_{H_2^{}}^{} = 2n_{O_2^{}}^{} + n_{Cl_2^{}}^{} = 0,03\;{\text{mol}}\)

\( \to {\text{Phan ung (1) va (2) la 2 phan ung vua du}}\)

\(C\% _{HCl}^{} = \frac{{0,06.36,5}}{{0,3.18 + 0,06.36,5}}.100 = 28,85\% \)

\(2.\)

\({\text{Trich mau thu cho moi lan thi nghiem}}\)

\({\text{lay cac mau thu tien hanh dun nong}}\)

\({\text{neu xuat hien khi' va ket tua thi do la Mg(HCO}}_3^{})_2^{}{\text{va Ba(HCO}}_3^{})_2^{}(N\hom 1)\)

\(Mg(HCO_3^{})_2^{} \to MgCO_3^{} + H_2^{}O + CO_2^{}\)

\(Ba(HCO_3^{})_2^{} \to BaCO_3^{} + H_2^{}O + CO_2^{}\)

\({\text{Neu chi xuat hien khi' thi do la KHCO}}_3^{}\)

\(KHCO_3^{} \to K_2^{}CO_3^{} + CO_2^{} + H_2^{}O\)

\({\text{Khong co hien tuong la NaHSO}}_4^{}{\text{ va Na}}_2^{}CO_3^{}(N\hom 2)\)

\(Cho\;(1){\text{ vao (2)}}\)

\({\text{neu thi nghiem vua co khi va ket tua thi do la Ba(HCO}}_3^{})_2^{}\;{\text{o nhom' 1 va NaHSO}}_4^{}\;{\text{o nhom 2}}\)

\({\text{Ba(HCO}}_3^{})_2^{} + NaHSO_4^{} \to BaSO_4^{} + Na_2^{}SO_4^{} + H_2^{}O + CO_2^{}\)