Sơ đồ phản ứng xảy ra như sau:

 ${\text{NH}}_3\stackrel{+{\text{O}}_2,\text{Pt}}{\to }\text{NO}\to {\text{NO}}_2\to {\text{HNO}}_3$

Dung dịch \({\rm{HN}}{{\rm{O}}_3}\) có \({\rm{pH}} = 1 \Rightarrow \left[ {{{\rm{H}}^ + }} \right] = {10^{ - {\rm{pH}}}} = 0,1{\rm{M}} \Rightarrow {n_{{\rm{HN}}{{\rm{O}}_3}}} = 0,1.0,3 = 0,03\;{\rm{mol}}\)

Dựa trên sơ đồ phản ứng hoặc bảo toàn nguyên tố (N) dễ thấy \({n_{{\rm{HN}}{{\rm{O}}_3}}} = {n_{{\rm{N}}{{\rm{H}}_3}}} = 0,03\;{\rm{mol}}\)

 $\begin{array}{l}4{\text{NH}}_3+5{\text{O}}_2\stackrel{Pt,t^{°}}{\to }4\text{NO}+6{\text{H}}_2\text{O}\\ 0,03 0,0375 0,03mol\end{array}$

\(\begin{array}{l}2{\rm{NO}} + {{\rm{O}}_2} \to 2{\rm{N}}{{\rm{O}}_2}\\0,03\,\,\,\,\,\,0,015\,\,\,\,\,0,03\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol\end{array}\)

\(\begin{array}{l}4{\rm{N}}{{\rm{O}}_2} + {{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}} \to 4{\rm{HN}}{{\rm{O}}_3}\\0,03\,\,\,\,\,\,0,0075\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol\end{array}\)

\({n_{{O_2}}}_{bd} = \frac{V}{{24,79}} - {n_{N{H_3}}} = \frac{V}{{24,79}} - 0,03\)mol

\({n_{{O_2}({\rm{sau}}\,{\rm{pu}})}} = \frac{V}{{24,79}} - 0,03 - 0,0375 - 0,015 - 0,0075 = \frac{V}{{4.24,79}}\)

\( \Rightarrow {\rm{V}} = 2,9748 \approx 2,975\)lít.

Chọn B.