Đáp án:
b) \({V_{kk}} = 65,74{\text{ lít}}\)
c) \({m_{C{O_2}}} = 17,217{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng:
\({C_2}{H_5}OH + 3{O_2}\xrightarrow{{}}2C{O_2} + 3{H_2}O\)
Ta có:
\({V_{{C_2}{H_5}OH}} = 25.45\% = 11,25ml \to {m_{{C_2}{H_5}OH}} = 11,25.0,8 = 9{\text{ gam}} \to {{\text{n}}_{{C_2}{H_5}OH}} = \frac{9}{{46}}{\text{ mol}}\)
Ta có:
\({n_{{O_2}}} = 3{n_{{C_2}{H_5}OH}} = \frac{9}{{46}}.3 = \frac{{27}}{{46}}{\text{ mol}} \to {V_{{O_2}}} = \frac{{27}}{{46}}.22,4 = 13,148{\text{ lít}} \to {{\text{V}}_{kk}} = 13,148.5 = 65,74{\text{ lít}}\)
\({n_{C{O_2}}} = 2{n_{{C_2}{H_5}OH}} = \frac{9}{{46}}.2 = \frac{9}{{23}}{\text{ mol}} \to {m_{C{O_2}}} = \frac{9}{{23}}.44 = 17,217{\text{ gam}}\)
