Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
a,
\(\begin{array}{l}
a,\\
{N_2} + {O_2} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} 2NO\\
NO + \dfrac{1}{2}{O_2} \to N{O_2}\\
4N{O_2} + {O_2} + 2{H_2}O \to 4HN{O_3}\\
F{e_2}{O_3} + 6HN{O_3} \to 2Fe{(N{O_3})_3} + 3{H_2}O\\
4Fe{(N{O_3})_3} \to 2F{e_2}{O_3} + 12N{O_2} + 3{O_2}\\
b,\\
2N{H_4}N{O_3} \to 2{N_2} + 4{H_2}O + {O_2}\\
{N_2} + 2{O_2} \to 2N{O_2}\\
{N_2} + 3{H_2} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} 2N{H_3}\\
CuC{l_2} + 2N{H_3} + 2{H_2}O \to Cu{(OH)_2} + 2N{H_4}Cl\\
2NaOH + 2N{O_2} \to NaN{O_2} + NaN{O_3} + {H_2}O\\
2NaN{O_3} \to 2NaN{O_2} + {O_2}\\
c,\\
4N{H_3} + 5{O_2} \to 4NO + 6{H_2}O\\
NO + \dfrac{1}{2}{O_2} \to N{O_2}\\
4N{O_2} + {O_2} + 2{H_2}O \to 4HN{O_3}\\
P + 5HN{O_3} \to 5N{O_2} + {H_3}P{O_4} + {H_2}O\\
3CaO + 2{H_3}P{O_4} \to C{a_3}{(P{O_4})_2} + 3{H_2}O\\
C{a_3}{(P{O_4})_2} + 3N{a_2}C{O_3} \to 2N{a_3}P{O_4} + 3CaC{O_3}\\
d,\\
{N_2} + 3{H_2} \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} 2N{H_3}\\
4N{H_3} + 5{O_2} \to 4NO + 6{H_2}O\\
NO + \dfrac{1}{2}{O_2} \to N{O_2}\\
4N{O_2} + {O_2} + 2{H_2}O \to 4HN{O_3}\\
CuO + 2HN{O_3} \to Cu{(N{O_3})_2} + {H_2}O\\
2Cu{(N{O_3})_2} \to 2CuO + 4N{O_2} + {O_2}\\
CuO + {H_2} \to Cu + {H_2}O\\
Cu + C{l_2} \to CuC{l_2}\\
CuC{l_2} + 2NaOH \to 2NaCl + Cu{(OH)_2}
\end{array}\)
