1)
PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
Theo đề bài, \(Na_2O\) phản ứng hết
\(\Rightarrow n_{Na_2O}=\frac{12,4}{62}=0,2\left(mol\right)\) \(\Rightarrow n_{NaOH}=0,4\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,4\cdot40=16\left(g\right)\)
Ta có: \(m_{dd}=m_{Na_2O}+m_{H_2O}=12,4+150=162,4\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\frac{16}{162,4}\cdot100\approx9,85\%\)
2)
Ta có: \(n_{NaOH\left(3M\right)}=3\cdot6=18\left(mol\right)\)
\(\Rightarrow V_{NaOH\left(2M\right)}=\frac{18}{0,2}=90\left(l\right)\)
\(\Rightarrow\) Cần thêm \(90-6=84\left(l\right)\)