M_X = 21,8.2 = 43,6; n _X = 0,25 mol

⇒ m_X = 0,25.43,6 = 10,9g

n_CO2 = 3n_X = 0,75 ⇒ mCO₂ = 0,75.44 = 33g

m_H = m_X – m_C = 10,9 – 12.0,75 = 1,9 ⇒ n_H = 1,9 ⇒ n_H2O = 0,9 mol

 ⇒ m_H2O = 17,1g

Đáp án A.

Bài 1:

\(M_X=21,8.M_{H_2}=21,8.2=43,6\left(\dfrac{g}{mol}\right)\\ Đặt:a=n_{C_3H_8}\left(mol\right)\left(a>0\right)\\ \Rightarrow\dfrac{44.a+42.\left(0,25-a\right)}{0,25}=43,6\\ \Leftrightarrow a=0,2\left(mol\right)\\ \Rightarrow n_{C_3H_8}=0,2\left(mol\right);n_{C_3H_6}=0,05\left(mol\right)\\ PTHH:C_3H_8+5O_2\rightarrow\left(t^o\right)3CO_2+4H_2O\\ 2C_3H_6+9O_2\rightarrow\left(t^o\right)6CO_2+6H_2O\\ n_{CO_2}=3.n_{hh}=3.0,25=0,75\left(mol\right)\\ \Rightarrow m_{CO_2}=0,75.44=33\left(g\right)\\ n_{H_2O}=4.0,2+3.0,05=0,95\left(mol\right)\\ \Rightarrow m_{H_2O}=18.0,95=17,1\left(g\right)\)

Bài 2:

\(n_{C_2H_5OH}=\dfrac{230}{46}=5\left(mol\right)\\ C_2H_4+H_2O\rightarrow\left(H^+,t^o\right)C_2H_5OH\\ n_{C_2H_4\left(LT\right)}=n_{C_2H_5OH}=5\left(mol\right)\\ Vì:H=40\%\Rightarrow n_{C_2H_4\left(TT\right)}=5:40\%=12,5\left(mol\right)\\ \Rightarrow m_{C_2H_4\left(TT\right)}=28.12,5=350\left(g\right)\)

Bài 3:

\(V_{khí.bay.ra}=V_{CH_4}=2,688\left(l\right)\\ \Rightarrow V_{anken}=5,6-2,688=2,912\left(l\right)\\ \Rightarrow n_{anken}=\dfrac{2,912}{22,4}=0,13\left(mol\right)\\ m_{bình.tăng}=m_{anken}=7,28\left(g\right)\\ \Rightarrow M_{Anken}=14n=\dfrac{7,28}{0,13}=56\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow n=4\\ \Rightarrow CTPT.anken:C_4H_8\)

Đáp án C

Chọn D.

\(n_X=\dfrac{0,896}{22,4}=0,08\left(mol\right)\)

\(M_X=21.2=42\left(g\text{/}mol\right)\\ \rightarrow m_X=0,08.42=3,36\left(g\right)\)

PTHH: 

\(C_3H_4+4O_2\xrightarrow[]{t^o}3CO_2+H_2O\\ 2C_3H_6+9O_2\xrightarrow[]{t^o}6CO_2+6H_2O\\ C_3H_8+5O_2\xrightarrow[]{t^o}3CO_2+4H_2O\)

Theo PTHH: \(n_C=n_{CO_2}=3n_X=3.0,08=0,24\left(mol\right)\)

\(\rightarrow V_{CO_2}=0,24.22,4=5,376\left(l\right)\)

BTNT: 

\(m_H=m_X=m_C=3,36-0,24.12=0,48\left(g\right)\\ \rightarrow n_H=\dfrac{0,48}{1}=0,48\left(mol\right)\)

Theo PTHH: \(n_{H_2O}=\dfrac{1}{2}n_H=\dfrac{1}{2}.0,48=0,24\left(mol\right)\)

\(\rightarrow m_{H_2O}=0,24.18=3,42\left(g\right)\)

nX=0,89622,4=0,08(mol)nX=0,89622,4=0,08(mol)

${}^{}=21.2=42\left(g\text{/}mol\right)\to {}^{}=0,08.42=3,36\left(g\right)$

PTHH: 

${}^{}{}^{}+4{}^{}{}^{}3C{}^{}+{}^{}O2{}^{}{}^{}+9{}^{}{}^{}6C{}^{}+6{}^{}O{}^{}$

Gọi công thức hóa học chung của hỗn hợp X là \(C_3H_x\)

Có \(\overline{M_X}=21\cdot2=42đvC\Rightarrow12\cdot3+x\cdot1=42\)

\(\Rightarrow x=6\)

\(n_X=\dfrac{1,12}{22,4}=0,05mol\)

\(C_3H_6+\dfrac{9}{2}O_2\rightarrow3CO_2+3H_2O\)

0,05                     0,15        0,15

\(m_{CO_2}=0,15\cdot44=6,6g\)

\(m_{H_2O}=0,15\cdot18=2,7g\)

\(C_3H_4+4O_2\underrightarrow{^{^{t^0}}}3CO_2+2H_2O\)

\(C_3H_6+\dfrac{9}{2}O_2\underrightarrow{^{^{t^0}}}3CO_2+3H_2O\)

\(C_3H_8+5O_2\underrightarrow{^{^{t^0}}}3CO_2+4H_2O\)

\(n_X=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(\Rightarrow n_{CO_2}=3\cdot n_X=3\cdot0.05=0.15\left(mol\right)\)

\(m_{CO_2}=0.15\cdot44=6.6\left(g\right)\)

\(m_X=21\cdot2\cdot0.05=2.1\left(g\right)\)

\(\Rightarrow m_H=m_X-m_C=2.1-0.15\cdot12=0.3\left(g\right)\)

\(n_H=0.3=0.3\left(mol\right)\)

\(\Rightarrow n_{H_2O}=0.15\left(mol\right)\)

\(m_{H_2O}=0.15\cdot18=2.7\left(g\right)\)

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