\(n_{CO_2}=\dfrac{17,6}{44}=0,4\left(mol\right)\)

=> n_C = 0,4 (mol)

\(n_{H_2O}=\dfrac{8,1}{18}=0,45\left(mol\right)\)

=> n_H = 0,9 (mol)

\(n_{N_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

=> n_N = 0,1 (mol)

\(n_O=\dfrac{10,3-0,4.12-0,9.1-0,1.14}{16}=0,2\left(mol\right)\)

Xét n_C : n_H : n_O : n_N = 0,4 : 0,9 : 0,2 : 0,1 = 4 : 9 : 2 : 1

=> CTPT: (C₄H₉O₂N)_n

Mà M_X = 51,5.2 = 103 g/mol

=> n = 1

=> CTPT: C₄H₉O₂N

D

chắc D

bn tham khảo:

C₂H₅Cl+NaOH->C₂H₅OH+NaCl

C₂H₅OH-to>C₂H₄+H₂O

C₂H₄+HBr->C₂H₅Br

C₂H₅Br+KOH->KBr+H₂O+C₂H₄

C₂H₄+H₂-to>C₂H₆

C₂H₆-craking->C₂H₄+H₂

C₂H₄+3O₂-to>2CO₂+2H₂O

CO₂+Ca(OH)₂->CaCO₃+H₂O

Câu 1:

a. \(C_2H_4+H_2O\)  \(\underrightarrow{t^o,H_2SO_4}\)  \(C_2H_5OH\)

b. \(nCH_2=CH_2\)  \(\underrightarrow{t^o,xt}\)  \(\left(-CH_2-CH_2-\right)_n\)

c. \(CH_2=CH-CH=CH_2+2H_2\)  \(\underrightarrow{t^o,Ni}\)  \(CH_3-CH_2-CH_2-CH_3\)

d. \(2C_3H_6+9O_2\rightarrow6CO_2+6H_2O\)

2. 

3.

Đặt công thức tổng quát hỗn hợp X là \(C_3H_x\)

Ta có: \(M_X=12\times3+x=42\Leftrightarrow x=6\Rightarrow C_3H_6\) 

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

0.05                       0.15       0.15

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

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

\(m=m_{CO_2}=0.15\times44=6.6g\)

Vậy bình nước vôi trong tăng 6.6g

C₂H₄+Br₂->C₂H₄Br₂

0,15--0,15 mol

n _Br2=\(\dfrac{24}{160}\)=0,15 mol

=>V_C2H4=0,15.22,4=3,36l

=>V_CH4=4,48-3,36=1,12l

ta có :

\(\left\{{}\begin{matrix}M_{hhkhí}=0,6.29=17,4\\n_{hhkhí}=\dfrac{3,36}{22,4}=0,15\end{matrix}\right.\)

=>PP đừng chéo : \(\dfrac{n_{CH4}}{n_{C2H4}}\)=\(\dfrac{53}{7}\)

=>\(\left\{{}\begin{matrix}nCH4=0,1325\\nC2H4=0,0175\end{matrix}\right.\)

=>BTNT:

Σ n _CaCO3=n _CH4+2n_C2H2=0,1675 mol

=>m _CaCO3=0,1675.100=16,75g

bn tham khảo:

chữ xấu quá

nhh=\(\dfrac{1680}{22400}\)=0,075 mol

n hidrocacbono =\(\dfrac{1120}{22400}\)=0,05 mol

=> n hidrocacbon ko no =\(\dfrac{1680-1120}{22400}\)=0,025 mol

Đặt các công thức là  : C_nH_2n+2-2kBr_2k

C_nH_2n+2-2k+kBr₂-> C_nH_2n+2-2kBr_2k

0,025----------0,025k

0,025k=\(\dfrac{4}{160}\)

=>k=1

=>Vậy hidrocacbon là C_nH_2n

C_mH_2n+2+\(\dfrac{3m+1}{2}\)O₂-to>m CO₂+ (m+1)H₂O

0,05---------0,05m

C_nH_2n+\(\dfrac{3n}{2}\)O₂-to>n CO₂+n H₂O

0,025---0,025n

CO₂+Ca(OH)₂->CaCO₃+H₂O

                           0,125 mol

0,05m+0,025n=0,125 mol

2m+n=5

n         2     3     4

m       1,5    1    0,5

n=3->m=1 (nhận )

=>A=C₃H₆, CH₄