Đáp án:

Giải thích các bước giải:

6 gam kết tủa nc2ag2=0.025mol

=>nc2h2=0.025mol

Vc2h2=0.56 lit  =>%c2h2=25%

vc2h4=1.68 lít =>%c2h4=75%

Em tham khảo nha:

\(\begin{array}{l}
6)\\
{C_2}{H_2} + 2AgN{O_3} + 2N{H_3} \to {C_2}A{g_2} + 2N{H_4}N{O_3}\\
{n_{{C_2}A{g_2}}} = \dfrac{6}{{240}} = 0,025\,mol\\
{n_{{C_2}{H_2}}} = {n_{{C_2}A{g_2}}} = 0,025\,mol\\
\% {V_{{C_2}{H_2}}} = \dfrac{{0,025 \times 22,4}}{{2,24}} \times 100\%  = 25\% \\
\% {V_{{C_2}{H_4}}} = 100 - 25 = 75\% \\
7)\\
a)\\
\text{ Gọi a là số mol của $H_2O$ }\\
{n_{C{O_2}}} = {n_{{H_2}O}} = a\,mol\\
 \Rightarrow {m_{C{O_2}}} - {m_{{H_2}O}} = 39g\\
 \Rightarrow 44a - 18a = 39 \Rightarrow a = 1,5\,mol\\
BTNT\,O:\\
2{n_{{O_2}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}} \Rightarrow {n_{{O_2}}} = 2,25\,mol\\
{V_{{O_2}}} = 2,25 \times 22,4 = 50,4l\\
b)\\
{n_{hh}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
\overline n  = \dfrac{{{n_{C{O_2}}}}}{{{n_{hh}}}} = \dfrac{{1,5}}{{0,4}} = 3,75\\
 \Rightarrow hh:{C_3}{H_6};{C_4}{H_8}\\
hh:{C_3}{H_6}(a\,mol),{C_4}{H_8}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,4\\
3a + 4b = 1,5
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,3\\
\% {V_{{C_3}{H_6}}} = \dfrac{{0,1}}{{0,4}} \times 100\%  = 25\% \\
\% {V_{{C_4}{H_8}}} = 100 - 25 = 75\% 
\end{array}\)