Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\) => 27a + 56b = 21,9 (*)
\(n_{O_2}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\)
PTHH:
\(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)
a----->0,75a--->0,5a
\(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
b----->\(\dfrac{2}{3}b\)---->\(\dfrac{1}{3}b\)
=> \(0,75a+\dfrac{2}{3}b=0,475\) (**)
Từ (*), (**) => \(\left\{{}\begin{matrix}a=0,5\\b=0,15\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,5.27}{21,9}.100\%=61,64\%\\\%m_{Fe}=100\%-61,64\%=38,36\%\end{matrix}\right.\)
mchất rắn sau phản ứng = 21,9 + 0,475.32 = 37,1 (g)
=> \(\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{0,5.102.0,5}{37,1}.100\%=68,73\%\\\%m_{Fe_3O_4}=100\%-68,73\%=31,27\%\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Fe}=y\end{matrix}\right.\) ( mol )
\(\rightarrow m_{hh}=27x+56y=21,9\left(g\right)\) (1)
\(n_{O_2}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\)
\(4Al+3O_2\rightarrow\left(t^o\right)Al_2O_3\)
`x` `3/4 x` `1/4 x` ( mol )
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
`y` `2/3 y` `1/3 y` ( mol )
\(\rightarrow n_{O_2}=\dfrac{3}{4}x+\dfrac{2}{3}y=0,475\left(mol\right)\) (2)
\(\left(1\right);\left(2\right)\rightarrow\left\{{}\begin{matrix}x=0,5\\y=0,15\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,5.27}{21,9}.100=61,64\%\\\%m_{Fe}=100\%-61,64\%=38,36\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}m_{Al_2O_3}=\dfrac{1}{4}.0,5.102=12,75\left(g\right)\\m_{Fe_3O_4}=\dfrac{1}{3}.0,15.232=11,6\left(g\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{12,75}{12,75+11,6}.100=52,36\%\\\%m_{Fe_3O_4}=100\%-52,36\%=47,64\%\end{matrix}\right.\)
