Đáp án:
\(\% {m_{CaC{O_3}}} = 64,3\%; \% {m_{CaC{l_2}}} = 35,7\% \)
\(C{\% _{CaC{l_2}}} = 27,22\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(CaC{O_3} + HCl\xrightarrow{{}}CaC{l_2} + C{O_2} + {H_2}O\)
Ta có:
\({n_{C{O_2}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol = }}{{\text{n}}_{CaC{O_3}}}\)
\( \to {m_{CaC{O_3}}} = 0,2.100 = 20{\text{ gam}} \to {{\text{m}}_{CaC{l_2}}} = 31,1 - 20 = 11,1{\text{ gam}}\)
\( \to {n_{CaC{l_2}}} = \frac{{11,1}}{{40 + 35,5.2}} = 0,1{\text{ mol}}\)
\(\% {m_{CaC{O_3}}} = \frac{{20}}{{31,1}}.100\% = 64,3\% \to \% {m_{CaC{l_2}}} = 35,7\% \)
\({n_{CaC{l_2}}} = {n_{CaC{O_3}}} + {n_{CaC{l_2}}} = 0,2 + 0,1 = 0,3{\text{ mol}}\)
\( \to {m_{CaC{l_2}}} = 0,3.(40 + 35,5.2) = 33,3{\text{ gam}}\)
\({n_{HCl}} = 2{n_{C{O_2}}} = 0,4{\text{ mol}} \to {{\text{m}}_{HCl}} = 0,4.36,5 = 14,6{\text{ gam}}\)
\( \to {m_{dd\;{\text{HCl}}}} = \frac{{14,6}}{{14,6\% }} = 100{\text{ gam}}\)
BTKL:
\({m_A} + {m_{dd\;{\text{HCl}}}} = {m_{dd}} + {m_{C{O_2}}}\)
\( \to 31,1 + 100 = {m_{dd}} + 0,2.44 \to {m_{dd}} = 122,3{\text{ gam}}\)
\( \to C{\% _{CaC{l_2}}} = \frac{{33,3}}{{122,3}}.100\% = 27,22\% \)
