Đáp án:
\(\begin{array}{l}
a)\\
{m_{BaS{O_4}}} = 23,3g\\
b)\\
C{\% _{HCl}} = 1,5\% \\
C{\% _{{H_2}S{O_4}}} = 2,6\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{H_2}S{O_4} + BaC{l_2} \to 2HCl + BaS{O_4}\\
{m_{BaC{l_2}}} = \dfrac{{400 \times 5,2}}{{100}} = 20,8g\\
{n_{BaC{l_2}}} = \dfrac{{20,8}}{{208}} = 0,1mol\\
{m_{dd{H_2}S{O_4}}} = 100 \times 1,14 = 114g\\
{m_{{H_2}S{O_4}}} = \dfrac{{114 \times 20}}{{100}} = 22,8g\\
{n_{{H_2}S{O_4}}} = \dfrac{{22,8}}{{98}} = 0,23mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,23}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,1mol\\
{m_{BaS{O_4}}} = 0,1 \times 233 = 23,3g\\
b)\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 0,2mol\\
{m_{HCl}} = 0,2 \times 36,5 = 7,3g\\
{n_{{H_2}S{O_4}_d}} = 0,23 - 0,1 = 0,13mol\\
{m_{{H_2}S{O_4}}} = 0,13 \times 98 = 12,74g\\
{m_{ddspu}} = 400 + 114 - 23,3 = 490,7g\\
C{\% _{HCl}} = \dfrac{{7,3}}{{490,7}} \times 100\% = 1,5\% \\
C{\% _{{H_2}S{O_4}}} = \dfrac{{12,74}}{{490,7}} \times 100\% = 2,6\%
\end{array}\)
