\(2NaOH\left(0,02006\right)+H_2SO_4\left(0,01003\right)\rightarrow Na_2SO_4\left(0,01003\right)+2H_2O\)
\(m_{NaOH}=25.0,04=1\left(g\right)\)
\(\Rightarrow n_{NaOH}=\dfrac{1}{40}=0,025\left(mol\right)\)
\(V_{H_2SO_4}=\dfrac{51}{1,02}=50\left(ml\right)=0,05\left(l\right)\)
\(\Rightarrow n_{H_2SO_4}=0,05.0,2=0,01\left(mol\right)\)
Theo đề bài thì NaOH dư 0,26%
\(\Rightarrow m_{NaOH\left(d\text{ư}\right)}=0,26\%.\left(25+51\right)=0,1976\left(g\right)\)
\(\Rightarrow n_{NaOH\left(d\text{ư}\right)}=\dfrac{0,1976}{40}=0,00494\left(mol\right)\)
\(\Rightarrow n_{NaOH\left(p\text{ứ}\right)}=0,025-0,00494=0,02006\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4\left(d\text{ư}\right)}=0,05-0,01003=0,03997\left(mol\right)\\n_{Na_2SO_4}=0,01003\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4\left(d\text{ư}\right)}=0,03997.98=3,91706\left(g\right)\\m_{Na_2SO_4}=0,01003.142=1,42426\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C\%\left(NaOH\right)=0,26\%\\C\%\left(H_2SO_4\right)=\dfrac{3,91706}{25+51}.100\%=5,15\%\\C\%\left(Na_2SO_4\right)=\dfrac{1,42426}{25+51}.100\%=1,87\%\end{matrix}\right.\)
