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a)$CH_4 + Cl_2 \xrightarrow{ánh\ sáng} CH_3Cl + HCl$
b) $2KHCO_3 \xrightarrow{t^o} K_2CO_3 + CO_2 + H_2O$
c) $C_2H_2 + \dfrac{5}{2}O_2 \xrightarrow{t^o} 2CO_2 + H_2O$
d) $C_2H_6 + Cl_2 \xrightarrow{ánh\ sáng} C_2H_5Cl + HCl$
e) $Ca(HCO_3)_2 \xrightarrow{t^o} CaCO_3 + CO_2 + H_2O$
f) $CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
\(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,1<--0,1
=> \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1.22,4}{5,6}.100\%=40\%\\\%V_{CH_4}=100\%-40\%=60\%\end{matrix}\right.\)
a) Khí thoát ra là CH4
\(n_{CH_4}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
\(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{3,7185}{14,874}.100\%=25\%\\\%V_{C_2H_4}=100\%-25\%=75\%\end{matrix}\right.\)
b)
\(n_{C_2H_4}=\dfrac{14,874.75\%}{24,79}=0,45\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,45-->0,45
=> \(C_{M\left(dd.Br_2\right)}=\dfrac{0,45}{0,15}=3M\)
c)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,15---------------------->0,3
C2H4 + 3O2 --to--> 2CO2 + 2H2O
0,45------------------------->0,9
=> mH2O = (0,3 + 0,9).18 = 21,6 (g)
- \(nCH_2=CH_2\underrightarrow{t^o,p,xt}\left(-CH_2-CH_2-\right)_n\)
- \(CH_4+Cl_2\underrightarrow{as}CH_3Cl+HCl\)
- \(CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)
- \(CH\equiv CH+2Br_2\rightarrow CHBr_2-CHBr_2\)
- \(C_6H_{12}O_6+Ag_2O\underrightarrow{NH_3}C_6H_{12}O_7+2Ag\)
a, Khí tác dụng với dd Brom: C2H4.
b, Ta có: \(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,05\left(mol\right)\Rightarrow m_{C_2H_4}=0,05.28=1,4\left(g\right)\)
a) C2H4 + Br2 --> C2H4Br2
b) nBr2 = 0,2.0,2 = 0,04 (mol)
PTHH: C2H4 + Br2 --> C2H4Br2
0,04<--0,04
=> \(m_{C_2H_4}=0,04.28=1,12\left(g\right)\)
\(m_{CH_4}=n_{CH_4}.M_{CH_4}=\left(\dfrac{1,12}{22,4}-0,04\right).16=0,16\left(g\right)\)
c) \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,04.22,4}{1,12}.100\%=80\%\\\%V_{CH_4}=100\%-80\%=20\%\end{matrix}\right.\)
a) \(CH_4+Cl_2\underrightarrow{as}CH_3Cl+HCl\) (pư thế)
b) \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\) (pư cộng)
Em cảm ơn ạ