ta có :

nBr2=\(\dfrac{16}{160}=0,1mol\)

C2H4+Br2->C2H4Br2

0,1------0,1

=>VC2H4=0,1.22,4=2,24l

=>VCH4=3,36l->n CH4=0,15 mol

->%VC2H4=\(\dfrac{2,24}{5,6}.100\)=40%

=>%VCH4=60%

c)

CH4+2O2-to>CO2+2H2O

0,15---------------0,15

C2H4+3O2--to>2CO2+2H2O

0,1--------------------0,2
=>m CaCO3=0,35.100=35g

a, Ta có: \(n_{CH_4}+n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(1\right)\)

\(16n_{CH_4}+28n_{C_2H_4}=3\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,1\left(mol\right)\\n_{C_2H_4}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,1.16}{3}.100\%\approx53,33\%\\\%m_{C_2H_4}\approx46,67\%\end{matrix}\right.\)

- Ở cùng điều kiện nhiệt độ và áp suất, % số mol cũng là %V.

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,15}.100\%\approx66,67\%\\\%V_{C_2H_4}\approx33,33\%\end{matrix}\right.\)

b, PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Có: m _tăng = m_C2H4 = 0,05.28 = 1,4 (g)

a) \(n_{hh}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=0,15\\16a+28b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,15}.100\%=66,67\%\\\%V_{C_2H_4}=100\%-66,67\%=33,33\%\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,1.16}{3}.100\%=53,33\%\\\%m_{C_2H_4}=100\%-53,33\%=46,67\%\end{matrix}\right.\)

b) \(m=m_{C_2H_4}=0,05.28=1,4\left(g\right)\)

a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Ta có: \(n_{C_2H_4}+n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(1\right)\)

Theo PT: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=\dfrac{48}{160}=0,3\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,2.22,4}{5,6}.100\%=80\%\\\%V_{C_2H_2}=20\%\end{matrix}\right.\)

b, \(V_{ddBr_2}=\dfrac{0,3}{0,25}=1,2\left(M\right)\)

a) 

\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

           0,05<--0,05

=> \(V_{C_2H_4}=0,05.22,4=1,12\left(l\right)\)

=> \(V_{CH_4}=4,48-1,12=3,36\left(l\right)\)

b) \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{1,12}{4,48}.100\%=25\%\\\%V_{CH_4}=\dfrac{3,36}{4,48}.100\%=75\%\end{matrix}\right.\)

`C_2H_4+Br_2->C_2H_4Br_2`

`0,09---0,09mol`

`n(Br_2)=(14,4)/160=0,09mol`

`->%(C_2H_4)=((0,09.22,4)/(3,36)).100%=60%`

`->%(CH_4)=100-60=40%`

#yBTrCute

C2,H2 hay C2H2?

C2H2+2Br2->C2H2Br4

0,05-----0,1

n Br2=\(\dfrac{16}{160}=0,1mol\)

=>%VC2H2=\(\dfrac{0,05.22,4}{3,36}100\)=33,3%

=>%VCH4=66,7%

Gọi số mol của \(C_2H_2\)  và \(CH_4\)  lần lượt là x và y.

\(C_2H_2+2Br_2\rightarrow CHBr_2+CHBr_2\)

 x           2x

\(CH_4+Br_2\rightarrow CH_3Br+HBr\)

  y         y 

Ta có hệ pt :

\(\left\{{}\begin{matrix}x+y=\dfrac{3,36}{22,4}\\2x+y=\dfrac{16}{160}\end{matrix}\right.\)

Giải hệ ta được : x = -0,05:))

coi lại đề mỗi cái đề cx đưa ko đàng hoàng nx.-.

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

b) Ta có: \(n_{Br_2}=\dfrac{11,2}{160}=0,07\left(mol\right)=n_{C_2H_4}\)

\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,07\cdot22,4}{1,72}\cdot100\%\approx91,16\%\) 

\(\Rightarrow\%V_{CH_4}=8,84\%\)

a) 

C₂H₄ + Br₂ --> C₂H₄Br₂

C₂H₂ + 2Br₂ --> C₂H₂Br₄

b) Gọi số mol C₂H₄, C₂H₂ là a, b (mol)

=> \(a+b=\dfrac{11,2}{22,4}=0,5\) (1)

PTHH: C₂H₄ + Br₂ --> C₂H₄Br₂

                a---->a

             C₂H₂ + 2Br₂ --> C₂H₂Br₄

                b---->2b

=> a + 2b = \(\dfrac{112}{160}=0,7\) (2)

(1)(2) => a = 0,3 (mol); b = 0,2 (mol)

\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,3}{0,5}.100\%=60\%\\\%V_{C_2H_2}=\dfrac{0,2}{0,5}.100\%=40\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{C_2H_4}=\dfrac{0,3.28}{0,3.28+0,2.26}.100\%=61,765\%\\\%m_{C_2H_2}=\dfrac{0,2.26}{0,3.28+0,2.26}.100\%=38,235\%\end{matrix}\right.\)