Đáp án:
\(\begin{array}{l}
a)\\
{m_{N{a_2}S{O_4}}} = 14,2g\\
b)\\
{C_{{M_{N{a_2}S{O_4}}}}} = 0,5M\\
{C_{{M_{{H_2}S{O_4}}}}} = 1,5M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{8}{{40}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = {C_M} \times V = 0,2 \times 2 = 0,4mol\\
\dfrac{{0,2}}{2} < \dfrac{{0,4}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{N{a_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,1mol\\
{m_{N{a_2}S{O_4}}} = n \times M = 0,1 \times 142 = 14,2g\\
b)\\
{n_{{H_2}S{O_4}d}} = {n_{{H_2}S{O_4}}} - \dfrac{{{n_{NaOH}}}}{2} = 0,3mol\\
{C_{{M_{N{a_2}S{O_4}}}}} = \dfrac{n}{V} = \dfrac{{0,1}}{{0,2}} = 0,5M\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{n}{V} = \dfrac{{0,3}}{{0,2}} = 1,5M
\end{array}\)
