Giải thích các bước giải:
1.
PT ion rút gọn:
\(\begin{array}{l}
a.M{g^{2 + }} + 2O{H^ - } \to Mg{(OH)_2}\\
b.C{O_3}^{2 - } + 2{H^ + } \to C{O_2} + {H_2}O\\
c.HC{O_3}^ - + O{H^ - } \to C{O_3}^{2 - } + {H_2}O\\
d.A{l^{3 + }} + 3N{H_3} + 3{H_2}O \to Al{(OH)_3} + 3N{H_4}^ + \\
e.N{H_4}^ + + O{H^ - } \to N{H_3} + {H_2}O
\end{array}\)
\(\begin{array}{l}
1.\\
a.MgC{l_2} + NaOH \to Mg{(OH)_2} + NaCl\\
b.N{a_2}C{O_3} + {H_2}S{O_4} \to N{a_2}S{O_4} + C{O_2} + {H_2}O\\
c.NaOH + NaHC{O_3} \to N{a_2}C{O_3} + {H_2}O\\
d.AlC{l_3} + 3{H_2}O + 3N{H_3} \to Al{(OH)_3} + 3N{H_4}Cl\\
e.KOH + N{H_4}Cl \to KCl + N{H_3} + {H_2}O\\
2.\\
a.\\
Fe + 6HN{O_3} \to Fe{(N{O_3})_3} + 3N{O_2} + 3{H_2}O\\
A{l_2}{O_3} + 6HN{O_3} \to 2Al{(N{O_3})_3} + 3{H_2}O\\
b.\\
{n_{N{O_2}}} = 0,01mol\\
\to {n_{Fe}} = \dfrac{1}{3}{n_{Fe}} = 0,003mol \to {m_{Fe}} = 0,168g\\
\to {m_{Al}} = 1,412g \to {n_{Al}} = 0,014mol\\
\to \% {m_{Fe}} = \dfrac{{0,168}}{{1,58}} \times 100\% = 10,63\% \\
\to \% {m_{Al}} = 100\% - 10,63\% = 89,37\% \\
c.\\
{n_{HN{O_3}}} = 6{n_{Fe}} + 6{n_{A{l_2}{O_3}}} = 0,102mol\\
\to {V_{HN{O_3}}} = \dfrac{{0,102}}{1} + 20\% = 0,302l
\end{array}\)
