Đáp án:
\(\begin{array}{l}
a)\\
{m_{Al}} = 19,8g\\
b)\\
{n_{HN{O_3}}} = 2,8\,mol
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{hh}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
hh:N{O_2}(a\,mol),{N_2}(b\,mol)\\
a + b = 0,4\\
46a + 28b = 14,8\\
\Rightarrow a = b = 0,2\,mol\\
Al \to A{l^{ + 3}} + 3e\\
{N^{ + 5}} + 1e \to N{O_2}\\
{N^{ + 5}} + 10e \to {N_2}\\
BT\,e:3{n_{Al}} = 10{n_{{N_2}}} + {n_{N{O_2}}} \Rightarrow {n_{Al}} = \dfrac{{11}}{{15}}\,mol\\
{m_{Al}} = \dfrac{{11}}{{15}} \times 27 = 19,8g\\
b)\\
{n_{HN{O_3}}} = 2{n_{N{O_2}}} + 12{n_{{N_2}}} = 2,8\,mol
\end{array}\)
