sục 3 chất khí qua dd Ca(OH)2

-CO2: xuất hiện kết tủa trắng

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

-CH4,C2H2: ko hiện tượng

Tiếp tục sục 2 chất khí qua dd Brom dư

-C2H2: dd Brom mất màu

\(C_2H_2+2Br_2\rightarrow C_2H_4Br_2\)

-CH4: ko hiện tượng

Sửa PTHH: \(C_2H_2+2Br_2\rightarrow C_2H_4Br_2\)

Thành: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

\(n_{Br_2}=\dfrac{4}{160}=0,025mol\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

0,025     0,025                     ( mol )

\(V_{hh}=\dfrac{2,8}{22,4}=0,125mol\)

\(\%V_{C_2H_4}=\dfrac{0,025}{0,125}.100=20\%\)

\(\%V_{CH_4}=100\%-20\%=80\%\)

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

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

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

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

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

Dài quá!!!

\(a,n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ Theo.pt:n_{C_2H_4}=n_{Br_2}=0,2\left(mol\right)\\ n_{hhkhi}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{CO_2}=0,3-0,2=0,1\left(mol\right)\\ m_{C_2H_4}=0,2.28=5,6\left(g\right)\\ m_{CO_2}=0,1.44=4,4\left(g\right)\\ b,C_{MddBr_2}=\dfrac{0,2}{0,5}=0,4M\)

ô

có cùng số ntu cacbon nhé mn giúp mình sớm trong đêm nay với ạ<3

Mg+2HCl-to>MgCl2+H2

0,25----0,5------0,25-----0,25

n Mg=\(\dfrac{6}{24}\)=0,25 mol

=>VH2=0,25.22,4=5,6l

m MgCl2=0,25.95=23,75g

m HCl=0,5.36,5=18,25 g

=>m dd Hcl=100g

=>C%=\(\dfrac{23,75}{6+100-0,5}.100=22,511\%\)

Zn+2CH3COOH->(CH3COO)2Zn+H2

0,25-------0,5------------------------------0,25

n CH3COOH=\(\dfrac{30}{60}=0,5mol\)

=>m Zn=0,25.65=16,25g

=>VH2=0,25.22,4=5,6l

CH3COOH+C2H5OH->CH3COOC2H5+H2O

 1-------------------------------------1

n CH3COOH=1 mol

n C2H5OH=2,17 mol

=>C2H5OH dư

=>m CH3COOC2H5=1.88=88g

=>H=\(\dfrac{55}{88}100=62,5\%\)

1.\(n_{CH_3COOH}=\dfrac{30}{60}=0,5mol\)

\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

0,25          0,5                                           0,25 ( mol )

\(m_{Zn}=0,25.65=16,25g\)

\(V_{H_2}=0,25.22,4=5,6l\)

2.\(n_{CH_3COOH}=\dfrac{60}{60}=1mol\)

\(n_{C_2H_5OH}=\dfrac{100}{46}=2,17mol\)

\(n_{CH_3COOC_2H_5}=\dfrac{55}{88}=0,625mol\)

\(CH_3COOH+C_2H_5OH\rightarrow CH_3COOC_2H_5+H_2O\)

          1                 2,17                   0,625                        ( mol )

         0,625                                       0,625                      ( mol ) 

=> H tính théo CH3COOH

\(H=\dfrac{0,625}{1}.100=62,5\%\)

??

Gọi công thức phân tử của A là: CxHyOz

Ta có

x:y:z=\(\dfrac{18}{12}:\dfrac{3}{1}:\dfrac{8}{16}\)=1,5:3:0,5=>CTHH :C3H6O