Đáp án:
\({C_{{M_{{H_2}S{O_4}}}}} = 1,5M\)
Giải thích các bước giải:
\(\begin{array}{l}
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O(1)\\
N{a_2}C{O_3} + {H_2}S{O_4} \to N{a_2}S{O_4} + C{O_2} + {H_2}O(2)\\
{n_{C{O_2}}} = \dfrac{{2,8}}{{22,4}} = 0,125mol\\
{n_{{H_2}S{O_4}(2)}} = {n_{C{O_2}}} = 0,125mol\\
{n_{NaOH}} = 0,025 \times 2 = 0,05mol\\
{n_{{H_2}S{O_4}(1)}} = \dfrac{{{n_{NaOH}}}}{2} = 0,025mol\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}S{O_4}(1)}} + {n_{{H_2}S{O_4}(2)}} = 0,025 + 0,125 = 0,15mol\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{{0,15}}{{0,1}} = 1,5M
\end{array}\)
