Đáp án:
\(\begin{array}{l}
a)\\
{m_{BaS{O_4}}} = 23,3g\\
b)\\
{C_\% }{H_2}S{O_4} = 2,6\% \\
{C_\% }HCl = 1,49\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
{n_{BaC{l_2}}} = \dfrac{{400 \times 5,2\% }}{{208}} = 0,1\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{114 \times 20\% }}{{98}} \approx 0,23\,mol\\
{n_{BaC{l_2}}} < {n_{{H_2}S{O_4}}} \Rightarrow {H_2}S{O_4} \text{ dư }\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,1\,mol\\
{m_{BaS{O_4}}} = 0,1 \times 233 = 23,3g\\
b)\\
{n_{{H_2}S{O_4}}} \text{ dư }= 0,23 - 0,1 = 0,13\,mol\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 0,2\,mol\\
{m_{{\rm{dd}}spu}} = 114 + 400 - 23,3 = 490,7g\\
{C_\% }{H_2}S{O_4} \text{ dư }= \dfrac{{0,13 \times 98}}{{490,7}} \times 100\% = 2,6\% \\
{C_\% }HCl = \dfrac{{0,2 \times 36,5}}{{490,7}} \times 100\% = 1,49\%
\end{array}\)
