Đáp án:
\({m_{CaC{O_3}}}= 6{\text{ gam;}}{{\text{m}}_{CaO}} = 10,08{\text{ gam}}\)
\(H = 75\% \)
\( C{\% _{CaC{l_2}}} = 19,25\% ;C{\% _{HCl}} = 0,5273\% \)
Giải thích các bước giải:
Vì cho \(A\) tác dụng với \(HCl\) tạo khí \(CO_2\) nên \(A\) chứa \(CaCO_3\) dư.
Phản ứng xảy ra:
\(CaC{O_3}\xrightarrow{{{t^o}}}CaO + C{O_2}\)
\(CaC{O_3} + 2HCl\xrightarrow{{}}CaC{l_2} + C{O_2} + {H_2}O\)
\(CaO + 2HCl\xrightarrow{{}}CaC{l_2} + {H_2}O\)
Ta có:
\({n_{CaC{O_3}}} = \frac{{24}}{{100}} = 0,24{\text{ mol = }}{{\text{n}}_{CaO}} + {n_{CaC{O_3}{\text{ dư}}}}\)
\({m_{HCl}} = 125.14,6\% = 18,25{\text{ gam}}\)
\( \to {n_{HCl}} = \frac{{18,25}}{{36,5}} = 0,5{\text{ mol}}\)
\( \to {n_{HCl}} > 2{n_{CaO}} + 2{n_{CaC{O_3}}}\)
Vậy \(HCl\) dư
Ta có:
\({n_{C{O_2}}} = \frac{{1,344}}{{22,4}} = 0,06{\text{ mol = }}{{\text{n}}_{CaC{O_3}}} \to {n_{CaO}} = 0,24 - 0,06 = 0,18{\text{ mol}}\)
Trong \(A\):
\({m_{CaC{O_3}}} = 0,06.100 = 6{\text{ gam;}}{{\text{m}}_{CaO}} = 0,18.56 = 10,08{\text{ gam}}\)
\({n_{CaC{O_3}{\text{ phản ứng}}}} = {n_{CaO}} = 0,18{\text{ mol}}\)
Hiệu suất:
\(H = \frac{{0,18}}{{0,24}} = 75\% \)
\({n_{CaC{l_2}}} = {n_{CaC{O_3}}} = 0,24{\text{ mol}}\)
\( \to {m_{CaC{l_2}}} = 0,24.(40 + 35,5.2) = 26,64{\text{ gam}}\)
\({n_{HCl{\text{ dư}}}} = 0,5 - 0,24.2 = 0,02{\text{ mol}}\)
\( \to {m_{HCl{\text{ dư}}}} = 0,02.36,5 = 0,73{\text{ gam}}\)
BTKL:
\({m_A} + {m_{dd\;{\text{HCl}}}} = {m_B} + {m_{C{O_2}}}\)
\( \to 6 + 10,08 + 125 = {m_B} + 0,06.44\)
\( \to {m_B} = 138,44{\text{ gam}}\)
\( \to C{\% _{CaC{l_2}}} = \frac{{26,64}}{{138,44}} = 19,25\% ;C{\% _{HCl}} = \frac{{0,73}}{{138,44}} = 0,5273\% \)
