Đáp án:
\(\begin{array}{l}
b)\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 250g\\
c)\\
{C_\% }N{a_2}S{O_4} = 7,89\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
b)\\
{m_{NaOH}} = 200 \times 10\% = 20g\\
{n_{NaOH}} = \dfrac{{20}}{{40}} = 0,5\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,5}}{2} = 0,25\,mol\\
{m_{{H_2}S{O_4}}} = 0,25 \times 98 = 24,5g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{24,5}}{{9,8\% }} = 250g\\
c)\\
{m_{{\rm{dd}}spu}} = 200 + 250 = 450g\\
{n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,25\,mol\\
{C_\% }N{a_2}S{O_4} = \dfrac{{0,25 \times 142}}{{450}} \times 100\% = 7,89\%
\end{array}\)
