Đáp án:
\(C{\% _{HCl}} = 14,6\% \)
Giải thích các bước giải:
Ta có:
\({n_{HCl}} = \frac{{8,96}}{{22,4}} = 0,4{\text{ mol}}\)
\( \to {m_{HCl}} = 0,4.36,5 = 14,6{\text{ gam}}\)
BTKL:
\({m_{dd{\text{ A}}}} = {m_{HCl}} + {m_{{H_2}O}} = 14,6 + 85,4 = 100{\text{ gam}}\)
\( \to C{\% _{HCl}} = \frac{{{m_{HCl}}}}{{{m_{dd{\text{ A}}}}}} = \frac{{14,6}}{{100}} = 14,6\% \)
100 gam dung dịch \(A\) chứa 0,4 mol \(HCl\).
Suy ra 50 gam dung dịch \(A\) chứa 0,2 mol \(HCl\).
Cho 8 gam \(CaCO_3\) tác dụng với 50 gam dung dịch \(A\).
\(CaC{O_3} + 2HCl\xrightarrow{{}}CaC{l_2} + C{O_2} + {H_2}O\)
Ta có:
\({n_{CaC{O_3}}} = \frac{8}{{100}} = 0,08{\text{ mol < }}\frac{1}{2}{n_{HCl}}\)
\( \to {n_{CaC{l_2}}} = {n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,08{\text{ mol}}\)
\({n_{HCl{\text{ dư}}}} = 0,2 - 0,08.2 = 0,04{\text{ mol}}\)
BTKL:
\({m_{dd\;{\text{A}}}} + {m_{CaC{O_3}}} = {m_{dd{\text{ B}}}} + {m_{C{O_2}}}\)
\( \to 50 + 8 = {m_{dd{\text{B}}}} + 0,08.44 \to {m_{dd{\text{ B}}}} = 54,48{\text{ gam}}\)
\({m_{CaC{l_2}}} = 0,08.(40 + 35,5.2) = 8,88{\text{ gam}}\)
\({m_{HCl}} = 0,04.36,5 = 1,46{\text{ gam}}\)
\( \to C{\% _{CaC{l_2}}} = \frac{{8,88}}{{54,48}} = 16,3\% \)
\(C{\% _{HCl}} = \frac{{1,46}}{{54,48}} = 2,68\% \)
