Ta có: \(C_{\%_{NaOH}}=\dfrac{m_{NaOH}}{400}.100\%=4\%\)
=> \(m_{NaOH}=16\left(g\right)\)
=> \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)
PTHH: HCl + NaOH ---> NaCl + H2O
Theo PT: \(n_{NaCl}=n_{NaOH}=0,4\left(mol\right)\)
=> \(m_{NaCl}=0,4.58,5=23,4\left(g\right)\)