Đáp án:
\( \% {m_{CaC{O_3}}} = 78,125\% ; \% {m_{CaO}} = 21,875\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(CaO + 2HCl\xrightarrow{{}}CaC{l_2} + {H_2}O\)
\(CaC{O_3} + 2HCl\xrightarrow{{}}CaC{l_2} + C{O_2} + {H_2}O\)
Ta có:
\({n_{C{O_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol = }}{{\text{n}}_{CaC{O_3}}}\)
\( \to {m_{CaC{O_3}}} = 0,2.100 = 20{\text{ gam}} \to {{\text{m}}_{CaO}} = 5,6{\text{ gam}}\)
\( \to {n_{CaO}} = \frac{{5,6}}{{40 + 16}} = 0,1{\text{ mol}}\)
\( \to \% {m_{CaC{O_3}}} = \frac{{20}}{{25,6}} = 78,125\% \to \% {m_{CaO}} = 21,875\% \)
\({n_{HCl}} = 2{n_{CaC{O_3}}} + 2{n_{CaO}} = 0,2.2 + 0,1.2 = 0,6{\text{ mol}}\)
\( \to {m_{HCl}} = 0,6.36,5 = 21,9{\text{ gam}}\)
\( \to {m_{dd{\text{ HCl}}}} = \frac{{21,9}}{{14,6\% }} = 150{\text{ gam}}\)
BTKL:
\({m_A} + {m_{dd{\text{ HCl}}}} = {m_{dd{\text{ B}}}} + {m_{C{O_2}}}\)
\( \to 25,6 + 150 = {m_{dd\;{\text{B}}}} + 0,2.44 \to {m_{dd{\text{ B}}}} = 166,8{\text{ gam}}\)
\({n_{CaC{l_2}}} = {n_{CaO}} + {n_{CaC{O_3}}} = 0,1 + 0,2 = 0,3{\text{ mol}}\)
\( \to {m_{CaC{l_2}}} = 0,3.(40 + 35,5.2) = 33,3{\text{ gam}}\)
\( \to C{\% _{CaC{l_2}}} = \frac{{33,3}}{{166,8}} = 19,964\% \)
