1)
Ancol có dạng \(C_nH_{2n+1}OH\)
Phản ứng xảy ra:
\({C_n}{H_{2n + 1}}OH + 1,5n{O_2}\xrightarrow{{{t^o}}}nC{O_2} + (n + 1){H_2}O\)
Ta có:
\({n_{C{O_2}}} = \frac{{6,6}}{{44}} = 0,15{\text{ mol}}\)
\( \to {n_{{O_2}}} = 1,5{n_{C{O_2}}} = 0,15.1,5 = 0,225{\text{ mol}}\)
\( \to {V_{{O_2}}} = 0,225.22,4 = 5,04{\text{ lít}}\)
2)
Phản ứng xảy ra:
\(2{C_2}{H_5}OH + 2Na\xrightarrow{{}}{C_2}{H_5}ONa + {H_2}\)
Ta có:
\({n_{{C_2}{H_5}OH}} = \frac{{4,6}}{{46}} = 0,1{\text{ mol = }}{{\text{n}}_{{C_2}{H_5}ONa}}\)
\( \to {m_{{C_2}{H_5}ONa}} = 0,2.68 = 13,6{\text{ gam}}\)
3)
Phản ứng xảy ra:
\({C_6}{H_{12}}{O_6}\xrightarrow{{men}}2{C_2}{H_5}OH + 2C{O_2}\)
Ta có:
\({n_{{C_2}{H_5}OH}} = \frac{{4,6}}{{46}} = 0,1{\text{ mol}}\)
\( \to {n_{{C_6}{H_{12}}{O_6}{\text{ lt}}}} = \frac{1}{2}{n_{C{O_2}}} = 0,05{\text{ mol}}\)
\( \to {n_{{C_6}{H_{12}}{O_6}}} = \frac{{0,05}}{{80\% }} = 0,0625{\text{ mol}}\)
\( \to {m_{{C_6}{H_{12}}{O_6}}} = 0,0625.180 = 11,25{\text{ gam}}\)
4)
Thiếu đề
5)
Ta có:
\({n_{{C_3}{H_5}{{(OH)}_3}}} = \frac{{1,84}}{{92}} = 0,02{\text{ mol}}\)
Phản ứng xảy ra:
\(2{C_3}{H_5}{(OH)_3} + Cu{(OH)_2}\xrightarrow{{}}{[Cu{(OH)_2}O]_2}Cu + 2{H_2}O\)
Ta có:
\({n_{Cu{{(OH)}_2}}} = \frac{1}{2}{n_{{C_3}{H_5}{{(OH)}_3}}} = 0,01{\text{ mol}}\)
\( \to {m_{{C_3}{H_5}{{(OH)}_3}}} = 0,01.92 = 0,92{\text{ gam}}\)
