Giải thích các bước giải:
\(\begin{array}{l}
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{m_{{H_2}S{O_4}}} = \dfrac{{50 \times 30\% }}{{100\% }} = 15g\\
\to {n_{{H_2}S{O_4}}} = 0,15mol\\
{m_{NaOH}} = \dfrac{{200 \times 10\% }}{{100\% }} = 20g\\
\to {n_{NaOH}} = 0,5mol\\
\to {n_{NaOH}} > {n_{{H_2}S{O_4}}} \to {n_{NaOH}}dư\\
{n_{NaOH}}(pt) = 2{n_{{H_2}S{O_4}}} = 0,3mol\\
\to {n_{NaOH}}(dư) = 0,2mol\\
a)\\
{n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,15mol\\
\to {m_{N{a_2}S{O_4}}} = 21,3g\\
b)\\
{m_{NaOH}}(du) = 8g\\
{m_{{\rm{dd}}Y}} = {m_{{\rm{dd}}NaOH}} + {m_{{\rm{dd}}{H_2}S{O_4}}} = 200 + 50 = 250g\\
\to C{\% _{NaOH}}(dư) = \dfrac{8}{{250}} \times 100\% = 3,2\% \\
\to C{\% _{N{a_2}S{O_4}}} = \dfrac{{21,3}}{{250}} \times 100\% = 8,52\%
\end{array}\)
