Đáp án:
\({m_{sp}} = 110,7{\text{ gam}}\)
\( {V_{{O_2}}} = 63,14{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2{N_2} + 5{O_2}\xrightarrow{{{t^o}}}2{N_2}{O_5}\)
\(2{H_2} + {O_2}\xrightarrow{{{t^o}}}2{H_2}O\)
Ta có:
\({m_{{N_2}}} = 20,5.70\% = 14,35{\text{ gam}}\)
\({{\text{m}}_{{H_2}}} = 20,5 - 14,35 = 6,15{\text{ gam}}\)
\( \to {n_{{N_2}}} = \frac{{14,35}}{{28}} = 0,5125{\text{ mol}}\)
\({n_{{H_2}}} = \frac{{6,15}}{2} = 3,075{\text{ mol}}\)
\( \to {n_{{N_2}{O_5}}} = {n_{{N_2}}} = 0,5125{\text{ mol}}\)
\({n_{{H_2}O}} = {n_{{H_2}}} = 3,075{\text{ mol}}\)
\( \to {m_{sp}} = {m_{{N_2}{O_5}}} + {m_{{H_2}O}}\)
\( = 0,5125.(14.2 + 16.5) + 3,075.18 = 110,7{\text{ gam}}\)
\({n_{{O_2}}} = \frac{5}{2}{n_{{N_2}}} + \frac{1}{2}{n_{{H_2}}} = 2,81875{\text{ mol}}\)
\( \to {V_{{O_2}}} = 2,81875.22,4 = 63,14{\text{ lít}}\)
