Đáp án:
\({m_{BaS{O_4}}} = 23,3g\)
\(\begin{array}{l}
C{\% _{HCl}} = 1,5\% \\
C{\% _{{H_2}S{O_4}(dư)}} = 3\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 100 \times 1,14 = 114g\\
\to {m_{{H_2}S{O_4}}} = \dfrac{{114 \times 21,5\% }}{{100\% }} = 24,51g\\
\to {n_{{H_2}S{O_4}}} = 0,25mol\\
{m_{BaC{l_2}}} = \dfrac{{400 \times 5,2\% }}{{100\% }} = 20,8g\\
\to {n_{BaC{l_2}}} = 0,1mol\\
\to {n_{{H_2}S{O_4}}} > {n_{BaC{l_2}}} \to {n_{{H_2}S{O_4}}}du\\
\to {n_{{H_2}S{O_4}}}(pt) = {n_{BaC{l_2}}} = 0,1mol\\
\to {n_{{H_2}S{O_4}}}(dư) = 0,15mol\\
\to {m_{{H_2}S{O_4}}}(dư) = 14,7g\\
\to {n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,1mol\\
\to {m_{BaS{O_4}}} = 23,3g\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 0,2mol\\
\to {m_{HCl}} = 7,3g\\
\to {m_{{\rm{dd}}}} = {m_{{\rm{dd}}{H_2}S{O_4}}} + {m_{{\rm{dd}}BaC{l_2}}} - {m_{BaS{O_4}}} = 490,7g\\
\to C{\% _{HCl}} = \dfrac{{7,3}}{{490,7}} \times 100\% = 1,5\% \\
\to C{\% _{{H_2}S{O_4}(dư)}} = \dfrac{{14,7}}{{490,7}} \times 100\% = 3\%
\end{array}\)
Đề phải là 400g dd BaCl2 mới tính được nha.
