Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
a. PTHH: 2Na + 2H2O ---> 2NaOH + H2↑
b. Ta có: \(n_{H_2O}=\dfrac{97,8}{18}=5,43\left(mol\right)\)
Ta thấy: \(\dfrac{0,1}{2}< \dfrac{5,43}{2}\)
=> H2O dư.
Theo PT: \(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
=> \(V_{H_2}=0,05.22,4=1,12\left(lít\right)\)
c. Ta có: \(m_{dd_{NaOH}}=2,3+97,8=100,1\left(g\right)\)
Theo PT: \(n_{NaOH}=n_{Na}=0,1\left(mol\right)\)
=> \(m_{NaOH}=0,1.40=4\left(g\right)\)
=> \(C_{\%_{NaOH}}=\dfrac{4}{100,1}.100\%=3,996\%\)