Theo đề bài: \(m_{giảm}=m_{H_2O}=54\left(g\right)\) \(\Rightarrow n_{H_2O}=\dfrac{54}{18}=3\left(mol\right)\)
PTHH: \(Mg\left(OH\right)_2\xrightarrow[]{t^o}MgO+H_2O\)
a_______a______a (mol)
\(4Fe\left(OH\right)_2+O_2\xrightarrow[]{t^o}2Fe_2O_3+4H_2O\)
b_____________\(\dfrac{1}{2}\)b______b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}58a+90b=204\\a+b=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2,0625\\b=0,9375\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{MgO}=2,0625\cdot40=82,5\left(g\right)\\m_{Fe_2O_3}=160\cdot\dfrac{0,9375}{2}=75\left(g\right)\\\%m_{Mg\left(OH\right)_2}=\dfrac{2,0625\cdot58}{204}\cdot100\%\approx58,64\%\\\%m_{Fe\left(OH\right)_2}=41,36\%\end{matrix}\right.\)
