Đáp án:

\(\% {V_{{C_2}{H_2}}} =  66,67\%; \% {V_{C{H_4}}} = 33,33\% \)

\(\% {m_{{C_2}{H_2}}} = 76,47\% ;\% {m_{C{H_4}}} = 24,53\% \)

\({m_{PVC}} = 8,5{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\({C_2}{H_2} + 2B{r_2}\xrightarrow{{}}{C_2}{H_2}B{{\text{r}}_4}\)

Khối lượng dung dịch tăng là khối lượng của \(C_2H_2\)

\( \to {n_{{C_2}{H_2}}} = \frac{{5,2}}{{12.2 + 2}} = 0,2{\text{ mol}}\)

\( \to {V_{{C_2}{H_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)

\( \to \% {V_{{C_2}{H_2}}} = \frac{{4,48}}{{6,72}} = 66,67\%  \to \% {V_{C{H_4}}} = 33,33\% \)

\({n_{hh}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol}} \to {{\text{n}}_{C{H_4}}} = 0,1{\text{ mol}}\)

\( \to {m_{C{H_4}}} = 0,1.16 = 1,6{\text{ gam}}\)

\( \to \% {m_{{C_2}{H_2}}} = \frac{{5,2}}{{5,2 + 1,6}} = 76,47\%  \to \% {m_{C{H_4}}} = 24,53\% \)

Điều chế \(PVC\)

\({C_2}{H_2} + HCl\xrightarrow{{}}{C_2}{H_3}Cl\)

\(n{C_2}{H_3}Cl\xrightarrow{{{t^o},xt}}{( - C{H_2} - CHCl - )_n}\)

Ta có:

\({n_{{C_2}{H_2}}} = {n_{{C_2}{H_3}Cl{\text{ lt}}}}=0,2\)

\( \to {m_{{C_2}{H_3}Cl{\text{ lt}}}} = {m_{PVC{\text{ lt}}}} = 0,2.(12.2 + 3 + 35,5) = 12,5{\text{ gam}}\)

\( \to {m_{PVC}} = 12,5.68\%  = 8,5{\text{ gam}}\)

Đáp án:

\(\begin{array}{l} a)\\ C_2H_2+2Br_2\to C_2H_2Br_4\\b)\\\%V_{C_2H_2}=66,67\%,\ \%V_{CH_4}=33,33\%\\\%m_{C_2H_2}=76,47\%,\ \%m_{CH_4}=23,53\%\\c)\ m_{PVC}=8,5(g)\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a)\\ C_2H_2+2Br_2\to C_2H_2Br_4\\b)\\ m_{\rm dd\ brom\ tăng}=m_{C_2H_2}\\n_{C_2H_2}=\dfrac{5,2}{26}=0,2(mol)\\n_{hh}=\dfrac{6,72}{22,4}=0,3(mol)\\\to n_{CH_4}=0,3-0,2=0,1(mol)\\\%V_{C_2H_2}=\dfrac{0,2}{0,3}×100\%=66,67\%\\\%V_{CH_4}=100\%-66,67\%=33,33\%\\\%m_{C_2H_2}=\dfrac{0,2×26}{0,2×26+0,1×16}×100\%=76,47\%\\\%m_{CH_4}=100\%-76,47\%=23,53\%\\c)\\C_2H_2+HCl\xrightarrow{t^o,\ xt}C_2H_3Cl\\nCH_2=CHCl\xrightarrow{t^o,\ xt,\ p}(\kern-6pt- CH_2-CHCl -\kern-6pt)_n\\n_{C_2H_2}=n_{C_2H_3Cl}=0,2(mol)\\m_{C_2H_3Cl}=m_{PVC}=0,2×62,5×68\%=8,5(g)\end{array}\)