Đáp án:
Bạn tham khảo lời giải ở dưới nhé
Giải thích các bước giải:
\(\begin{array}{l}
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
{m_{{H_2}S{O_4}}} = \dfrac{{100 \times 19,6\% }}{{100\% }} = 19,6g\\
\to {n_{{H_2}S{O_4}}} = 0,2mol\\
{m_{BaC{l_2}}} = \dfrac{{400 \times 13\% }}{{100\% }} = 52g\\
\to {n_{BaC{l_2}}} = 0,25mol\\
\to {n_{{H_2}S{O_4}}} < {n_{BaC{l_2}}} \to {n_{BaC{l_2}}}dư\\
a)\\
{n_{BaS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,2mol\\
\to {m_{BaS{O_4}}} = 46,6g\\
b)\\
{n_{HCl}} = 2{n_{{H_2}S{O_4}}} = 0,4mol \to {m_{HCl}} = 14,6g\\
{n_{BaC{l_2}(pt)}} = {n_{{H_2}S{O_4}}} = 0,2mol\\
\to {n_{BaC{l_2}(dư)}} = 0,05mol \to {m_{BaC{l_2}(du)}} = 10,4g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}{H_2}S{O_4}}} + {m_{{\rm{dd}}BaC{l_2}}} - {m_{BaS{O_4}}} = 453,4g\\
\to C{\% _{BaC{l_2}(dư)}} = \dfrac{{10,4}}{{453,4}} \times 100\% = 2,29\% \\
\to C{\% _{HCl}} = \dfrac{{14,6}}{{453,4}} \times 100\% = 3,2\%
\end{array}\)
