`a,`
`n_{H_2}=\frac{0,112}{22,4}=0,005(mol)`
`2Na+2H_2O->2NaOH+H_2`
Theo phương trình
`n_{Na}=2n_{H_2}=0,01(mol)`
`->m_{Na}=0,01.23=0,23(g)`
`->%m_{Na}=\frac{0,23}{20}.100=1,15%`
`%m_{Na_2O}=100-1,15=98,85%`
`b,`
`n_{NaOH}=n_{Na}+2n_{Na_2O}`
`->n_{NaOH}=0,01+2.\frac{19,77}{62}=\frac{502}{775}(mol)`
`->m_{NaOH}\approx 25,91(g)`
`m_{dd}=20+150-0,01.2=169,98(g)`
`->C%_{NaOH}=\frac{25,91}{169,98}.100\approx 15,24%`