PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\) (1)
\(Na_2O+H_2O\rightarrow2NaOH\) (2)
a) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) \(\Rightarrow n_{Na}=0,2\left(mol\right)\)
\(\Rightarrow m_{Na}=0,2\cdot23=4,6\left(g\right)\) \(\Rightarrow m_{Na_2O}=6,2\left(g\right)\)
b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{NaOH\left(1\right)}=n_{Na}=0,2\left(mol\right)\\n_{NaOH\left(2\right)}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{NaOH}=0,4\left(mol\right)\) \(\Rightarrow m_{NaOH}=0,4\cdot40=16\left(g\right)\)
c) Theo các PTHH: \(n_{H_2O}=n_{Na}+n_{Na_2O}=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2O}=0,3\cdot18=5,4\left(g\right)\)
