Đáp án:
3)
\(\begin{array}{l}
\% {m_{Al}} = 49,1\% \\
\% {m_{Fe}} = 50,9\%
\end{array}\)
Giải thích các bước giải:
3)
\(\begin{array}{l}
Al + 4HN{O_3} \to Al{(N{O_3})_3} + NO + 2{H_2}O\\
Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + 2{H_2}O\\
{n_{NO}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
a + b = 0,3(1)\\
\text{Theo đề ta có:}\\
27a + 56b = 11(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,2;b = 0,1\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% {m_{Al}} = \dfrac{{5,4}}{{11}} \times 100\% = 49,1\% \\
\% {m_{Fe}} = 100 - 49,1 = 50,9\% \\
4)\\
2N{H_4}Cl + Ca{(OH)_2} \to CaC{l_2} + 2N{H_3} + 2{H_2}O\\
2N{H_3} + 3CuO \to 3Cu + 3{H_2}O + {N_2}\\
{N_2} + {O_2} \to NO\\
2NO + {O_2} \to 2N{O_2}\\
N{O_2} + {H_2}O \to HN{O_3} + NO\\
HN{O_3} + NaOH \to NaN{O_3} + {H_2}O\\
2NaN{O_3} \to 2NaN{O_2} + {O_2}
\end{array}\)
