PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=\dfrac{200\cdot4\%}{40}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{150\cdot9,8\%}{98}=0,15\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\) \(\Rightarrow\) Axit còn dư, tính theo Bazơ
\(\Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05\cdot98=4,9\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2SO_4}=\dfrac{0,1\cdot142}{200+150}\cdot100\%\approx4,06\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{4,9}{200+150}\cdot100\%=1,4\%\end{matrix}\right.\)
