Đáp án đúng: C
0,78.
$\displaystyle {{\text{m}}_{{{\text{O}}_{\text{2}}}}}\text{=12,48-10,08=2,4g }\Rightarrow \text{ }{{\text{n}}_{{{\text{O}}_{\text{2}}}}}\text{=0,075mol}$
$\begin{array}{l}\text{Fe}\xrightarrow{{}}\overset{\text{+2}}{\mathop{\text{Fe}}}\,\text{+2e}\\\text{0,18}\,\,\xrightarrow{{}}\,\,\,\,\text{0,36}\end{array}$
$\begin{array}{l}\,\,\,{{\text{O}}_{\text{2}}}\,\,\,\text{+}\,\,\,\,\,\,\text{4e}\xrightarrow{{}}\text{2}\overset{\text{-2}}{\mathop{\text{O}}}\,\\\text{0,075}\to \text{0,3}\end{array}$
$\begin{array}{l}\text{Cu}\xrightarrow{{}}\overset{\text{+2}}{\mathop{\text{Cu}}}\,\text{+2e}\\\text{0,15}\,\,\xrightarrow{{}}\,\,\,\,\,\,\,0,3\end{array}$
$\begin{array}{l}\overset{\text{+5}}{\mathop{\text{N}}}\,\text{+}\,\,\text{3e}\xrightarrow{{}}\overset{\text{+2}}{\mathop{\text{N}}}\,\\\text{x}\,\to \,\text{3x}\end{array}$
Bte: 0,36+0,30=0,3+3.x→ x=0,12
$\displaystyle \Rightarrow {{\text{n}}_{\text{HN}{{\text{O}}_{\text{3}}}}}\text{= 2}\text{.}{{\text{n}}_{\text{Fe}}}\text{+2}\text{.}{{\text{n}}_{\text{Cu}}}\text{+}{{\text{n}}_{\text{NO}}}\text{=0,78mol}$
