Đáp án:

\( {V_{HN{O_3}}}  = 0,63125{\text{ lít}}\)

Giải thích các bước giải:

Phản ứng xảy ra:

\(3Zn + 8HN{O_3}\xrightarrow{{}}3Zn{(N{O_3})_2} + 2NO + 4{H_2}O\)

\(F{e_2}{O_3} + 6HN{O_3}\xrightarrow{{}}Fe{(N{O_3})_3} + 3{H_2}O\)

Ta có:

\({n_{NO}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}}\)

\( \to {n_{Zn}} = \frac{3}{2}{n_{NO}} = 0,3{\text{ mol}}\)

\( \to {m_{Zn}} = 0,3.65 = 19,5{\text{ gam}} \to {{\text{m}}_{F{e_2}{O_3}}} = 65,5 - 19,5 = 46{\text{ gam}}\)

\( \to {n_{F{e_2}{O_3}}} = \frac{{46}}{{56.2 + 16.3}} = 0,2875{\text{ mol}}\)

\(\to {n_{HN{O_3}}} = \frac{8}{3}{n_{Zn}} + 6{n_{F{e_2}{O_3}}}\)

\( = 0,3.\frac{8}{3} + 6.0,2875 = 2,525{\text{ mol}}\)

\( \to {V_{HN{O_3}}} = \frac{{2,525}}{4} = 0,63125{\text{ lít}}\)