Đáp án:
\(\begin{array}{l}
{C_\% }HCl = 3,65\% \\
{C_\% }{H_2}S{O_4} = 9,8\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
NaOH + HCl \to NaCl + {H_2}O\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
hh:HCl(a\,mol),{H_2}S{O_4}(b\,mol)\\
{n_{NaOH}} = 0,3 \times 1 = 0,3\,mol\\
\left\{ \begin{array}{l}
a + 2b = 0,3\\
58,5a + 142b = 20,05
\end{array} \right.\\
\Rightarrow a = b = 0,1\,mol\\
{C_\% }HCl = \dfrac{{0,1 \times 36,5}}{{100}} \times 100\% = 3,65\% \\
{C_\% }{H_2}S{O_4} = \dfrac{{0,1 \times 98}}{{100}} \times 100\% = 9,8\%
\end{array}\)
