Đáp án:
\( {V_{{H_2}}} = 6,72{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + F{e_2}{O_3}\xrightarrow{{{t^o}}}A{l_2}{O_3} + 2Fe\)
BTKL:
\({m_{F{e_2}{O_3}}} + {m_{Al}} = {m_X}\)
\( \to 16 + m = 26,8 \to m = 10,8\)
\( \to {n_{F{e_2}{O_3}}} = \frac{{16}}{{56.2 + 16.3}} = 0,1{\text{ mol;}}{{\text{n}}_{Al}} = \frac{{10,8}}{{27}} = 0,4{\text{ mol > 2}}{{\text{n}}_{F{e_2}{O_3}}}\)
\( \to {n_{Al{\text{ dư}}}} = 0,4 - 0,1.2 = 0,2{\text{ mol}}\)
\(NaOH + Al + {H_2}O\xrightarrow{{}}NaAl{O_2} + \frac{3}{2}{H_2}\)
\( \to {n_{{H_2}}} = \frac{3}{2}{n_{Al}} = 0,3{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)
