Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
Câu 1:
\(\begin{array}{l}
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
NaOH + {H_2}S{O_4} \to NaHS{O_4} + {H_2}O\\
{n_{NaHS{O_4}}} = 0,03mol\\
{n_{N{a_2}S{O_4}}} = 0,02mol\\
\to {n_{{H_2}S{O_4}}} = {n_{NaHS{O_4}}} + {n_{N{a_2}S{O_4}}} = 0,05mol\\
\to {n_{NaOH}} = 2{n_{N{a_2}S{O_4}}} + {n_{NaHS{O_4}}} = 0,07mol\\
\to {m_{{H_2}S{O_4}}} = 4,9g\\
\to {m_{NaOH}} = 2,8g\\
\to a = {m_{{H_2}S{O_4}}}{\rm{dd}} = \dfrac{{4,9 \times 100}}{{24,5}} = 20g\\
\to b = {m_{NaOH}}{\rm{dd}} = \dfrac{{2,8 \times 100}}{8} = 35g
\end{array}\)
Dung dịch sau phản ứng gồm: \(N{a_2}S{O_4}\) và \(NaHS{O_4}\)
\(\begin{array}{l}
{m_{{\rm{dd}}}} = a + b = 55g\\
\to C{\% _{N{a_2}S{O_4}}} = \dfrac{{2,84}}{{55}} \times 100\% = 5,16\% \\
\to C{\% _{NaHS{O_4}}} = \dfrac{{3,6}}{{55}} \times 100\% = 6,55\%
\end{array}\)
Câu 2:
Giả sử có 100g dung dịch \({H_2}S{O_4}\)
\(\begin{array}{l}
{M_2}{O_3} + 3{H_2}S{O_4} \to {M_2}{(S{O_4})_3} + 3{H_2}O\\
{m_{{H_2}S{O_4}}} = \dfrac{{20 \times 100}}{{100}} = 20g\\
\to {n_{{H_2}S{O_4}}} = \dfrac{{20}}{{98}}mol\\
\to {n_{{M_2}{{(S{O_4})}_3}}} = \dfrac{1}{3}{n_{{H_2}S{O_4}}} = \dfrac{{10}}{{147}}mol\\
\to {m_{{M_2}{{(S{O_4})}_3}}} = \dfrac{{10}}{{147}} \times (2M + 288)\\
{n_{{M_2}{O_3}}} = \dfrac{1}{3}{n_{{H_2}S{O_4}}} = \dfrac{{10}}{{147}}mol\\
\to {m_{{M_2}{O_3}}} = \dfrac{{10}}{{147}} \times (2M + 48)\\
{m_{{\rm{dd}}}} = \dfrac{{10}}{{147}} \times (2M + 48) + 100\\
\to C{\% _{{M_2}{{(S{O_4})}_3}}} = \dfrac{{\dfrac{{10}}{{147}} \times (2M + 288)}}{{\dfrac{{10}}{{147}} \times (2M + 48) + 100}} \times 100 = 21,756\\
\to M = 27\\
\to A{l_2}{O_3}
\end{array}\)
