Đáp án đúng: B
13,92.
Sau phản ứng dung dịch chỉ chứa 1 muối⇒ Muối là CuSO4 với số mol là 0,01.2 = 0,02 mol
⇒ Số mol S là 0,02-0,01 = 0,01 mol
Ta có:
$\begin{array}{l}\text{C}{{\text{u}}_{\text{2}}}\text{S}\xrightarrow{{}}\overset{\text{+2}}{\mathop{\text{2Cu}}}\,\text{+}\overset{\text{+6}}{\mathop{\text{S}}}\,\text{+10e}\\\text{S}\xrightarrow{{}}\overset{\text{+6}}{\mathop{\text{S}}}\,\text{+6e}\\\overset{\text{+5}}{\mathop{\text{N}}}\,\text{+1e}\xrightarrow{{}}\overset{\text{+4}}{\mathop{\text{N}}}\,\end{array}$
$\text{BTe:}\,\,{{\text{n}}_{\text{N}{{\text{O}}_{\text{2}}}}}\text{=10}{{\text{n}}_{\text{C}{{\text{u}}_{\text{2}}}\text{S}}}\text{+6}{{\text{n}}_{\text{S}}}\text{=0,16}\,\text{mol}$
$\begin{array}{l}\,\,2\text{N}{{\text{O}}_{\text{2}}}\text{+}\,\text{2NaOH}\xrightarrow{{}}\text{NaN}{{\text{O}}_{\text{3}}}\text{+NaN}{{\text{O}}_{\text{2}}}\text{+}{{\text{H}}_{\text{2}}}\text{O}\\\text{0,16}\,\,\,\to \,\,\,\text{0,16}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\xrightarrow{{}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{0,08}\\{{\text{m}}_{\text{N}{{\text{O}}_{\text{2}}}}}\text{+}{{\text{m}}_{\text{NaOH}}}\text{=}\,{{\text{m}}_{\text{r}}}\text{+}{{\text{m}}_{{{\text{H}}_{\text{2}}}\text{O}}}\Rightarrow \,{{\text{m}}_{\text{r}}}\text{=13,92}\,\text{gam}\end{array}$
