Em tham khảo nha:
\(\begin{array}{l}
6)\\
a)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
hh:C{H_4}(a\,ml),{C_2}{H_4}(b\,ml)\\
\left\{ \begin{array}{l}
a + b = 36\\
2a + 3b = 97,2
\end{array} \right.\\
\Rightarrow a = 10,8;b = 25,2\\
\% {V_{C{H_4}}} = \dfrac{{10,8}}{{36}} \times 100\% = 30\% \\
\% {V_{{C_2}{H_4}}} = 100 - 30 = 70\% \\
b)\\
{V_{C{O_2}}} = 10,8 + 25,2 \times 2 = 61,2\,ml\\
7)\\
a)\\
C{H_3}COOH + {C_2}{H_5}OH \xrightarrow{t^0,xt} C{H_3}COO{C_2}{H_5} + {H_2}O\\
b)\\
{n_{C{H_3}COOH}} = \dfrac{{12}}{{60}} = 0,2\,mol\\
{n_{{C_2}{H_5}OH}} = \dfrac{{13,8}}{{46}} = 0,3\,mol\\
{n_{C{H_3}COOH}} < {n_{{C_2}{H_5}OH}} \Rightarrow \text{ $C_2H_5OH$ dư } \\
{n_{{C_2}{H_5}OH}} \text{ dư }= 0,3 - 0,2 = 0,1\,mol\\
{m_{{C_2}{H_5}OH}} \text{ dư }= 0,1 \times 46 = 4,6g\\
c)\\
{n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,2\,mol\\
H = \dfrac{{14,08}}{{0,2 \times 88}} \times 100\% = 80\% \\
d)\\
{V_{{C_2}{H_5}OH}} = \dfrac{{13,8}}{{0,8}} = 17,25\,ml\\
{D_r} = \dfrac{{17,25}}{{17,25 + 40,25}} \times 100 = {30^0}
\end{array}\)
