Giả sử có 1 lít hỗn hợp 3 dung dịch đã trộn.

\({\rm{N}}{{\rm{O}}_3}^ - + 4{{\rm{H}}^ + } + 3{\rm{e}} \to {\rm{NO}} + 2{{\rm{H}}_2}{\rm{O}}\)

A. \(\sum {{{\rm{n}}_{{{\rm{H}}^ + }}}} = 2 \cdot 1 + 1 = 3\;{\rm{mol}};{{\rm{n}}_{{\rm{N}}{{\rm{O}}_3}^ - }} = 1\;{\rm{mol}}\)

\( \Rightarrow {{\rm{n}}_{{\rm{NO sinh}}\,\,{\rm{ra }}}} = \frac{1}{4}{{\rm{n}}_{{{\rm{H}}^ + }}} = 0,75\;{\rm{mol}}{\rm{.}}\)

B. \(\sum {{n_{{{\rm{H}}^ + }}}} = 2 \cdot 1 + 1 = 3\;{\rm{mol}};{n_{N{O_3}^ - }} = 1\;{\rm{mol}}\)

\( \Rightarrow {{\rm{n}}_{{\rm{NO sinh}}\,\,{\rm{ra }}}} = \frac{1}{4}{{\rm{n}}_{{{\rm{H}}^ + }}} = 0,75\;{\rm{mol}}{\rm{.}}\)

C. \(\sum {{{\rm{n}}_{{{\rm{H}}^ + }}}} = 2 \cdot 1 + 1 + 1 = 4\;{\rm{mol}};{{\rm{n}}_{{\rm{N}}{{\rm{O}}_3}^ - }} = 1\;{\rm{mol}}\)

\( \Rightarrow {{\rm{n}}_{{\rm{NO sinh}}\,{\rm{ra }}}} = \frac{1}{4}{{\rm{n}}_{{{\rm{H}}^ + }}} = 1\;{\rm{mol}}{\rm{.}}\)

D. \(\sum {{{\rm{n}}_{{{\rm{H}}^ + }}}} = 1 + 1 = 2\;{\rm{mol}};{{\rm{n}}_{{\rm{N}}{{\rm{O}}_3}^ - }} = 2\;{\rm{mol}}\)

\( \Rightarrow {{\rm{n}}_{{\rm{NO sinh}}\,\,{\rm{ra}}}} = \frac{1}{4}{{\rm{n}}_{{{\rm{H}}^ + }}} = 0,5\;{\rm{mol}}{\rm{.}}\)

Chọn C.