\(a,n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\\ n_A=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\end{matrix}\right.\left(đk:a,b>0\right)\)
PTHH:
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)
0,2<---------------------0,2
CH4 + 2O2 --to--> CO2 + 2H2O
a--------------------->a
C2H4 + 3O2 --to--> 2CO2 + 2H2O
b----------------------->2b
b, => \(\left\{{}\begin{matrix}a+b=0,125\\a+2b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,05\left(mol\right)\\b=0,075\left(mol\right)\end{matrix}\right.\left(TM\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,05}{0,125}.100\%=40\%\\\%V_{C_2H_4}=100\%-40\%=60\%\end{matrix}\right.\)
