\(PTHH:Mg\left(OH\right)2\overrightarrow{to}MgO+H2O\)
...........................x............x..............x........(mol)
\(Fe\left(OH\right)3\overrightarrow{to}Fe2O3+3H2O\)
2x.......................x...................3x.......(mol)
\(x=n_{MgO}=n_{Fe2O3}\)
Ta có :
\(36=18x+3x.18\)
=>x = 0,5(mol)
\(m_{bđ}=0,5.58+0,5.107.2=136\left(g\right)\)