a)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
b)
\(n_{Br2}=\frac{80.5\%}{160}=0,025\left(mol\right)\)
\(n_{C2H4}=n_{Br2}=0,025\left(mol\right)\)
\(m_{CaH4}=0,025.28=0,7\left(g\right)\)
\(m_{CH4}=1-0,7=0,3\left(g\right)\)
c)
\(n_{CH4}=\frac{0,3}{16}=0,01875\left(mol\right)\)
\(CH_4+2O_2\rightarrow CO_2+2H_2O\)
\(C_2H_4+2O_2\rightarrow2CO_2+2H_2O\)
\(n_{CO2}=0,01875+0,025.2=0,06875\left(mol\right)\)
\(\Rightarrow V_{CO2}=0,06875.22,4=1,54\left(l\right)\)
a) CH4+Br2--->CH3Br+HBr
x----------x(mol)
C2H4+Br2---->C2H4Br2
y----y(mol)
Ta có
m Br2=80.5/100=4(g)
n Br2=4/160=0,025(mol)
Theo bài ra ta có hpt
\(\left\{{}\begin{matrix}16x+28y=1\\x+y=0,025\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\\\end{matrix}\right.\)
