Gọi số mol Fe(OH)3, Mg(OH)2 là a,b (mol)
PTHH: \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H2O\)
________a ------------> 0,5a____________(mol)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
b-----------------> b_________(mol)
=> \(\left\{{}\begin{matrix}107a+58b=27,2\\160.0,5a+40b=27,2-7,2\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,11\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{Fe\left(OH\right)_3}=0,2.107=21,4\\m_{Mg\left(OH\right)_2}=0,1.58=5,8\left(g\right)\\m_{Fe_2O_3}=160.0,5.0,2=16g\\m_{MgO}=40.0,1=4\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{Fe\left(OH\right)_3}=\frac{21,4}{27,2}.100\%=78,68\%\\\%m_{Mg\left(OH\right)_2}=\frac{5,8}{27,2}.100\%=21,32\%\end{matrix}\right.\)
