Đồng đẳng, đồng phân của ankan

BTKL : 

\(m_A=m_{C_3H_8}=8.8\left(g\right)\)

\(n_{C_3H_8}=\dfrac{8.8}{44}=0.2\left(mol\right)\)

\(\Rightarrow n_{C_3H_8\left(pư\right)}=0.2\cdot90\%=0.18\left(mol\right)\)

\(C_3H_8\underrightarrow{t^0,xt}C_aH_{2a+2}+C_bH_{2b}\left(a+b=3\right)\)

\(n_{C_3H_8\left(dư\right)}=0.02\left(mol\right)\)

\(M_A=\dfrac{8.8}{0.18\cdot2+0.02}=23.16\left(\dfrac{g}{mol}\right)\)

\(C_3H_8 \to H_2 + C_3H_6\\ C_3H_8 \to CH_4 + C_2H_4\\ C_3H_8 \to C_3H_{8\ dư}\\ m_A = m_{propan}= 8,8(gam)\\ n_{C_3H_8} = \dfrac{8,8}{44} = 0,2\\ n_{C_3H_8\ dư} = 0,2 - 0,2.90\% =0,02(mol)\\ n_A = 2n_{C_3H_8\ pư} + n_{C_3H_8\ dư} = 0,2.90\%.2 + 0,02=0,38(mol)\\ M_A = \dfrac{8,8}{0,38} = 23,167\)

\(C_4H_{10} \to C_2H_6 + C_2H_4\\ C_4H_{10} \to CH_4 + C_3H_6\\ C_4H_{10} \to C_4H_{10\ dư}\)

Theo PTHH, ta thấy : 

\(2V_{C_4H_{10}\ pư} + V_{C_4H_{10}\ dư} = 1010\\ V_{C_4H_{10}\ pư} + V_{C_4H_{10}\ dư} = 560\\ \Rightarrow V_{C_4H_{10}\ pư} = 450 ; V_{C_4H_{10}\ dư} = 110\\ H = \dfrac{450}{560}.100\% = 80,36\%\)

\(Coi m_{C_4H_{10}} = 58(gam)\\ m_{hidrocacbon} = m_{C_4H_{10}} = 58(gam)\\ n_{hidrocacbon} = \dfrac{58}{16,325.2} = 1,77(mol)\\ n_{C_4H_{10}\ pư} = 1,77 - 1 = 0,77(mol)\\ H = \dfrac{0,77}{1}.100\% = 77\%\)

1) C₅H₈

2) A

3) C

4) CH₄,C₂H₆,C₃H₈

5) A 

\(CH_3COONa + NaOH \xrightarrow{t^o,CaO} CH_4 + Na_2CO_3\)

1) C₅H₈

2) C₂H₂

3) 

4) 

CH₄ , C₂H₆ , C₃H₈

5) 

CH₃COONa 

Ta có: \(\left\{{}\begin{matrix}\overline{M}_{hhkhí}=0,6\cdot29=17,4\\n_{hhkhí}=\dfrac{3,36}{22.4}=0,15\left(mol\right)\end{matrix}\right.\)

Theo phương pháp đường chéo: \(\dfrac{n_{CH_4}}{n_{C_2H_4}}=\dfrac{53}{7}\) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,1325\left(mol\right)\\n_{C_2H_4}=0,0175\left(mol\right)\end{matrix}\right.\)

Bảo toàn nguyên tố: \(\Sigma n_{CaCO_3}=n_{CH_4}+2n_{C_2H_4}=0,1675\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0,1675\cdot100=16,75\left(g\right)\)

Ta có: \(\dfrac{n_t}{n_s}=\dfrac{M_s}{M_t}\Rightarrow\dfrac{1}{3}=\dfrac{12.2}{M_t}\Rightarrow M_t=72\)

Do đó X là $C_5H_{12}$

Khi mono clo hóa thu được $C_5H_{11}Cl$

$\Rightarrow \%m_{Cl}=33,3\%$