10)
Cho hỗn hợp tác dụng với dung dịch \(Br_2\)
\({C_2}{H_2} + 2B{r_2}\xrightarrow{{}}{C_2}{H_2}B{r_4}\)
Khí thoát ra là \(CH_4\) (metan)
\({n_{B{r_2}}} = \frac{{48}}{{80.2}} = 0,3{\text{ mol}}\)
\( \to {n_{{C_2}{H_2}}} = \frac{1}{2}{n_{B{r_2}}} = 0,15{\text{ mol}}\)
\( \to {V_{{C_2}{H_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)
\( \to V = {V_{C{H_4}}} = 8,96 - 3,36 = 5,6{\text{ lít}}\)
\(\% {V_{C{H_4}}} = \frac{{5,6}}{{8,96}} = 62,5\% \to \% {V_{{C_2}{H_2}}} = 37,5\% \)
Đốt cháy khí thoát ra
\(C{H_4} + 2{O_2}\xrightarrow{{{t^o}}}C{O_2} + 2{H_2}O\)
Ta có:
\( \to {V_{{O_2}{\text{ lt}}}} = 2{V_{C{H_4}}} = 5,6.2 = 11,2{\text{ lít}}\)
\( \to {V_{{O_2}}} = 11,2.75\% = 8,4{\text{ lít}}\)
\( \to {V_{kk}} = 5{V_{{O_2}}} = 8,4.5 = 42{\text{ lít}}\)
11)
Phản ứng xảy ra:
\(2C{H_3}COOH + Mg\xrightarrow{{}}{(C{H_3}COO)_2}Mg + {H_2}\)
Ta có:
\({n_{Mg}} = \frac{6}{{24}} = 0,25{\text{ mol}}\)
\( \to {n_{C{H_3}COOH}} = 2{n_{Mg}} = 0,25.2 = 0,5{\text{ mol}}\)
\( \to {V_{dd{\text{ C}}{{\text{H}}_3}COOH}} = \frac{{0,5}}{{0,5}} = 1{\text{ lít}}\)
\({n_{{H_2}}} = {n_{Mg}} = 0,25{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,25.22,4 = 5,6{\text{ lít}}\)
