Đáp án:
\({\text{C}}{{\text{\% }}_{HCl}} = 14,88\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\({H_2}S{O_4} + BaC{l_2}\xrightarrow{{}}BaS{O_4} + 2HCl\)
Ta có:
\({m_{BaC{l_2}}} = 400.5,2\% = 20,8{\text{ gam}} \to {{\text{n}}_{BaC{l_2}}} = \frac{{20,8}}{{137 + 35,5.2}} = 0,1{\text{ mol = }}{{\text{n}}_{{H_2}S{O_4}}} = {n_{BaS{O_4}}}\)
\({m_{{H_2}S{O_4}}} = 0,1.98 = 9,8{\text{ gam}} \to {\text{C}}{{\text{\% }}_{{H_2}S{O_4}}} = \frac{{9,8}}{{114}} = 8,6\% \)
\({m_B} = {m_{BaS{O_4}}} = 0,1.233 = 23,3{\text{ gam}}\)
BTKL: \({m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} + {m_{dd{\text{ BaC}}{{\text{l}}_2}}} = {m_{dd{\text{A}}}} + {m_B} \to {m_{dd\;{\text{A}}}} = 114 + 400 - 23,3 = 490,7{\text{ gam}}\)
Dung dịch A chỉ chứa HCl
\({n_{HCl}} = 2{n_{BaC{l_2}}} = 0,1.2 = 0,2{\text{ mol}} \to {{\text{m}}_{HCl}} = 0,2.36,5 = 7,3{\text{ gam}} \to {\text{C}}{{\text{\% }}_{HCl}} = \frac{{7,3}}{{490,7}} = 14,88\% \)
