PTHH: \(Na_2O+2HNO_3\rightarrow2NaNO_3+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{Na_2O}=\dfrac{24,8}{62}=0,4\left(mol\right)\\n_{HNO_3}=\dfrac{50,4}{63}=0,8\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,4}{1}=\dfrac{0,8}{2}\) \(\Rightarrow\) Cả 1 chất đều p/ứ hết
\(\Rightarrow n_{NaNO_3}=0,8\left(mol\right)\) \(\Rightarrow m_{NaNO_3}=0,8\cdot85=68\left(g\right)\)