a)

$V_{CH_4} = V_{khí\ thoát\ ra}  = 2,24(lít)$
$\%V_{CH_4} = \dfrac{2,24}{8,96}.100\% = 25\%$

$\%V_{C_2H_4} = 100\% -25\% = 75\%$

b)

$n_{Br_2} = n_{C_2H_4} = \dfrac{8,96.75\%}{22,4} = 0,3(mol)$
$C_{M_{Br_2}} = \dfrac{0,3}{0,2} = 1,5M$
$m_{tăng} = m_{C_2H_4} = 0,3.28 = 8,4(gam)$

Khí thoát ra là CH4 => V CH4= 2,24(lít)

%V CH4 = 2,24/6,72  .100% = 33,33%

n C2H4 = a(mol) ; n C2H2 b(mol)

=> a + b  = (6,72 - 2,24)/22,4 = 0,2(1)

m tăng = m C2H4 + m C2H2 = 28a + 26b = 5,4(2)

Từ (1)(2) suy ra a = b = 0,1

V C2H2 =  V C2H4 = 0,1.22,4 = 2,24(lít)

Vậy : 

%V C2H2 = %V C2H4 = (100%-33,33%)/2 = 33,335%

a) \(n_{Br_2\left(p\text{ư}\right)}=\dfrac{200.8\%}{160}=0,1\left(mol\right);n_{hh}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

             0,1<-----0,1

b) \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1}{0,3}.100\%=33,33\%\\\%V_{CH_4}=100\%-33,33\%=66,67\%\end{matrix}\right.\)

c) \(n_{CH_4}=0,3-0,1=0,2\left(mol\right)\)

PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)

             0,2--->0,4

            \(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)

              0,1--->0,3

\(\Rightarrow V_{O_2}=\left(0,3+0,4\right).24,79=17,353\left(l\right)\)

\(a,Gọi\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\\n_{C_2H_2}=c\left(mol\right)\end{matrix}\right.\\ n_{hhkhí}=0,4\left(mol\right)\\ n_{CO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\\ n_{Br_2}=\dfrac{64}{160}=0,4\left(mol\right)\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ Mol:a\rightarrow a\\ C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ Mol:b\rightarrow2b\\ CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ Mol:a\rightarrow2a\rightarrow a\)

\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Mol:b\rightarrow3b\rightarrow2b\\ 2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:c\rightarrow2,5c\rightarrow2c\\ Hệ.pt\left\{{}\begin{matrix}a+b+c=0,4\\b+2c=0,4\\a+2b+2c=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\\c=0,1\left(mol\right)\end{matrix}\right.\)

\(\%V_{CH_4}=\%V_{C_2H_2}=\dfrac{0,1}{0,4}=25\%\\ \%V_{C_2H_4}=\dfrac{0,2}{0,4}=50\%\)

\(m_{CH_4}=0,1.16=1,6\left(g\right)\\ m_{C_2H_4}=28.0,2=5,6\left(g\right)\\ m_{C_2H_2}=0,1.26=2,6\left(g\right)\\ \%m_{CH_4}=\dfrac{1,6}{1,6+5,6+2,6}=16,32\%\\ \%m_{C_2H_4}=\dfrac{5,6}{1,6+5,6+2,6}=57,14\%\\ \%m_{C_2H_2}=100\%-16,32\%-57,14\%=26,54\%\)

\(b,PTHH:C_2H_5OH\rightarrow C_2H_4+H_2O\\ Mol:0,2\leftarrow0,2\\ m_{C_2H_5OH}=0,2.46=9,2\left(g\right)\)

Dài quá!!!

\(n_{hhkhí\left(C_2H_4,CH_4\right)}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

\(m_{tăng}=m_{C_2H_4}=4,2\left(g\right)\\ n_{C_2H_4}=\dfrac{4,2}{28}=0,15\left(mol\right)\\ \%V_{C_2H_4}=\dfrac{0,15}{0,35}=42,85\%\\ \%V_{CH_4}=100\%-42,85\%=57,15\%\)

\(a,n_{hhkhí\left(C_2H_4,C_2H_2\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Br_2}=\dfrac{80}{160}=0,5\left(mol\right)\\ Gọi\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ Mol:a\rightarrow a\\ C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ Mol:b\rightarrow2b\\ Hệ.pt\left\{{}\begin{matrix}a+b=0,3\\a+2b=0,5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\\ \%V_{C_2H_4}=\dfrac{0,1}{0,3}=33,33\%\\ \%V_{C_2H_2}=100\%-33,335=66,67\%\)

\(b,PTHH:\\ C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Mol:0,1\rightarrow0,3\rightarrow0,2\\ 2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:0,2\rightarrow0,25\rightarrow0,4\\ n_{CO_2}=0,2+0,4=0,6\left(mol\right)\\ PTHH:Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ Mol:0,6\rightarrow0,6\rightarrow0,6\\ m_{CaCO_3}=0,6.100=60\left(g\right)\)

Đáp án C.

Đáp án C

Khí thoát ra khỏi bình là Y ; nCO₂ = 0,03 ; nH₂O= 0,04

→ Y là ankan → n_Y = 0,04 - 0,03 = 0,01 → Y là C₃H₈
Đốt cháy X thu được nCO₂ = nH₂O

→ nC₂H₂= nC₃H₈ = 0,01 mol
→ nC₂H₄ = ( 0,82 - 0,01 × 26 ) : 28 = 0,02 mol
→ n_X = 0,01 + 0,02 + 0,01 = 0,04 → V_X = 0,896 (l)

Khí thoát ra khỏi bình là Y; n(CO₂) = 0,03 mol; n(H₂O) = 0,04 mol

→ Y là ankan → n(Y) = 0,01 → Y là C₃H₈

Đốt cháy X thu được n(CO₂) = n(H₂O) → n(C₂H₂) = n(C₃H₈) = 0,01 mol

→ n(C₂H₄) = (0,82 – 0,01*26)/28 = 0,02 mol

→ n(X) = 0,01 + 0,02 + 0,01 = 0,04 → V(X) = 0,896 lít → Đáp án C