a)
Phần 1: Đặt \(\left\{{}\begin{matrix}n_{C_2H_5OH}=a\left(mol\right)\\n_{C_nH_m\left(OH\right)_3}=b\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\Rightarrow n_{OH}=2n_{H_2}=1,4\left(mol\right)\\ \Rightarrow a+3b=1,4\\ \Rightarrow a=1,4-3b\)
b) \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{35,84}{22,4}=1,6\left(mol\right)\\n_{H_2O}=2,2\left(mol\right)\end{matrix}\right.\)
BTNT C: \(n_{CO_2}=2n_{C_2H_5OH}+n.n_{C_nH_m\left(OH\right)_3}\)
\(\Rightarrow2a+bn=1,6\\ \Leftrightarrow2\left(1,4-3b\right)+bn=1,6\\ \Leftrightarrow\left(6-n\right)b=1,2\\ \Leftrightarrow b=\dfrac{1,2}{6-n}\left(1\right)\)
BTNT H: \(2n_{H_2O}=6n_{C_2H_5OH}+\left(m+3\right)n_{C_nH_m\left(OH\right)_3}\)
\(\Rightarrow6a+\left(m+3\right)b=4,4\)
\(\Leftrightarrow6\left(1,4-3b\right)+\left(m+3\right)b=4,4\)
\(\Leftrightarrow\left(15-m\right)b=4\\ \Leftrightarrow b=\dfrac{4}{15-m}\left(2\right)\)
Từ (1), (2) \(\Rightarrow\dfrac{1,2}{6-n}=\dfrac{4}{15-m}\)
\(\Leftrightarrow\dfrac{15-m}{6-n}=\dfrac{10}{3}\)
\(\Leftrightarrow3\left(15-m\right)=10\left(6-n\right)\)
\(\Leftrightarrow45-3m=60-10n\)
\(\Leftrightarrow10n-3m=15\)
\(\Rightarrow\left\{{}\begin{matrix}n=3\\m=5\end{matrix}\right.\)
Vậy \(C_nH_m\left(OH\right)_3:C_3H_5\left(OH\right)_3\)
b)
PTHH:
\(C_2H_5OH+3O_2\xrightarrow[]{t^o}2CO_2+3H_2O\)
a---------------------->2a------->3a
\(2C_3H_5\left(OH\right)_3+7O_2\xrightarrow[]{t^o}6CO_2+8H_2O\)
b--------------------------->3b------->4a
\(\Leftrightarrow\left\{{}\begin{matrix}2a+3b=1,6\\3a+4b=2,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,4\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{C_2H_5OH}=0,2.46=9,2\left(g\right)\\m_{C_3H_8\left(OH\right)_3}=0,4.92=36,8\left(g\right)\end{matrix}\right.\)