a) \(n_{hh}=\dfrac{4,48}{22,4}=0,2\left(mol\right);n_{CaCO_3}=\dfrac{56}{100}=0,56\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{C_2H_6}=a\left(mol\right)\\n_{C_4H_{10}}=b\left(mol\right)\end{matrix}\right.\Rightarrow a+b=0,2\left(1\right)\)
PTHH:
\(2C_2H_6+7O_2\xrightarrow[]{t^o}4CO_2+6H_2O\)
a------------------->2a
\(2C_4H_{10}+13O_2\xrightarrow[]{t^o}8CO_2+10H_2O\)
b---------------------->4b
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)
0,56<---0,56
\(\Rightarrow2a+4b=0,56\left(2\right)\)
Từ (1), (2) => a = 0,12; b = 0,08
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_6}=\%n_{C_2H_6}=\dfrac{0,12}{0,2}.100\%=60\%\\\%V_{C_4H_{10}}=100\%-60\%=40\%\end{matrix}\right.\)
Và \(\left\{{}\begin{matrix}\%m_{C_2H_6}=\dfrac{0,12.30}{0,12.30+0,08.58}.100\%=43,67\%\\\%m_{C_4H_{10}}=100\%-43,67\%=56,33\%\end{matrix}\right.\)
