PTHH: \(Na_2O+2HNO_3\rightarrow2NaNO_3+H_2O\)
Ta có: \(n_{Na_2O}=\frac{12,4}{62}=0,2\left(mol\right)\) \(\Rightarrow n_{HNO_3}=0,4mol\)
\(\Rightarrow\left\{{}\begin{matrix}m_{HNO_3}=0,4\cdot63=25,2\left(g\right)\\m_{NaNO_3}=27,25\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{HNO_3}=\frac{25,2}{400}\cdot100\%=6,3\%\\C\%_{NaNO_3}=\frac{27,25}{12,4+100}\cdot100\%\approx24,24\%\end{matrix}\right.\)