a) \(n_{Br_2\left(p\text{ư}\right)}=\dfrac{200.8\%}{160}=0,1\left(mol\right);n_{hh}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,1<-----0,1
b) \(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1}{0,3}.100\%=33,33\%\\\%V_{CH_4}=100\%-33,33\%=66,67\%\end{matrix}\right.\)
c) \(n_{CH_4}=0,3-0,1=0,2\left(mol\right)\)
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
0,2--->0,4
\(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)
0,1--->0,3
\(\Rightarrow V_{O_2}=\left(0,3+0,4\right).24,79=17,353\left(l\right)\)
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