Đáp án:
\(\begin{array}{l}
2.\\
a.Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + 2{H_2}O\\
b.{m_{Fe}} = 2,8g\\
c.{m_{F{e_2}{O_3}}} = 4g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1.\\
a.HN{O_3} + KOH \to KN{O_3} + {H_2}O\\
b.N{H_4}Cl + NaOH \to NaCl + N{H_3} + {H_2}O\\
c.N{a_2}C{O_3} + BaC{l_2} \to 2NaCl + BaC{O_3}\\
d.MgC{l_2} + 2NaOH \to 2NaCl + Mg{(OH)_2}\\
2.\\
a.Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + 2{H_2}O\\
b.\\
{n_{NO}} = 0,05mol\\
\to {n_{Fe}} = {n_{NO}} = 0,05mol\\
\to {m_{Fe}} = 2,8g\\
c.\\
4Fe{(N{O_3})_3} \to 2F{e_2}{O_3} + 12N{O_2} + 3{O_2}\\
{n_{Fe{{(N{O_3})}_3}}} = {n_{Fe}} = 0,05mol\\
\to {n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{Fe{{(N{O_3})}_3}}} = 0,025mol\\
\to {m_{F{e_2}{O_3}}} = 4g
\end{array}\)
