Đáp án:
Bạn tham khảo bài giải ở dưới nhé!!!
Giải thích các bước giải:
6,
\(\begin{array}{l}
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = 0,06mol\\
\to {n_{{H_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,03mol\\
\to {m_{{H_2}S{O_4}}} = 2,94g\\
\to {m_{{H_2}S{O_4}}}{\rm{dd}} = \dfrac{{2,94 \times 100}}{{30}} = 9,8g\\
\to {V_{{H_2}S{O_4}}}{\rm{dd}} = \dfrac{{{m_{{H_2}S{O_4}}}{\rm{dd}}}}{d} = 8,2ml\\
\to {n_{N{a_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,03mol\\
\to {m_{N{a_2}S{O_4}}} = 4,26g\\
\to {m_{NaOH}}{\rm{dd}} = 50 \times 1,047 = 52,35g\\
\to {m_{{\rm{dd}}}} = 9,8 + 52,35 = 62,15g\\
\to C{\% _{N{a_2}S{O_4}}} = \dfrac{{4,26}}{{62,15}} \times 100\% = 6,85\%
\end{array}\)
7,
\(\begin{array}{l}
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = \dfrac{{25 \times 4}}{{100 \times 40}} = 0,025mol\\
{V_{{H_2}S{O_4}}} = \dfrac{{{m_{dd}}}}{d} = 50ml\\
\to {n_{{H_2}S{O_4}}} = CM \times V = 0,01mol\\
\to \dfrac{{{n_{NaOH}}}}{2} > {n_{{H_2}S{O_4}}}
\end{array}\)
Suy ra dung dịch sau phản ứng có: NaOH dư và \(N{a_2}S{O_4}\)
\(\begin{array}{l}
\to {n_{NaOH}}dư= 0,025 - 2{n_{{H_2}S{O_4}}} = 0,005mol\\
\to {n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,01mol\\
\to C{\% _{N{a_2}S{O_4}}} = \dfrac{{0,01 \times 142}}{{25 + 51}} \times 100 = 1,87\% \\
\to C{\% _{NaOH}}dư= \dfrac{{0,005 \times 40}}{{25 + 51}} \times 100 = 0,26\%
\end{array}\)
