\(pthh:Na_2O+H_2O\rightarrow2NaOH\)
Mg+H2O -/->
NaOH - bazo kiềm
\(n_{NaOH}=0,05.2=0,1\left(mol\right)\)
\(\) \(n_{Na_2O}=\dfrac{1}{2}n_{NaOH}=0,05\left(mol\right)\\
m_{Na_2O}=0,05.62=3,1g\\
m_{MgO}=4,54-3,1=1,44g\\
m_{NaOH}=0,1.40=4g\\
m_{cr}=4+1,44=5,44g\)
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