Đáp án:
\( \% {m_{MgC{O_3}}} = 57,53\%; \% {m_{CuC{O_3}}} = 42,47\% \)
Giải thích các bước giải:
Gọi số mol \(MgCO_3;CuCO_3\) lần lượt là \(x;y\).
\( \to 84x + 124y = 14,6{\text{ gam}}\)
Ta có:
\({n_{C{O_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol = }}{{\text{n}}_{MgC{O_3}}} + {n_{CuC{O_3}}}\)
Ta có:
\({n_{C{O_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol = }}{{\text{n}}_{MgC{O_3}}} + {n_{CuC{O_3}}}\)
\( \to x + y = 0,15\)
Giải được:
\(x=0,1;y=0,05\)
\( \to {m_{MgC{O_3}}} = 0,1.84 = 8,4{\text{ gam}}\)
\( \to \% {m_{MgC{O_3}}} = \frac{{8,4}}{{14,6}} = 57,53\% \to \% {m_{CuC{O_3}}} = 42,47\% \)
BTKL:
\({n_{HCl}} = 2{n_{C{O_2}}} = 0,3{\text{ mol}}\)
\( \to {V_{dd{\text{ HCl}}}} = \frac{{0,3}}{2} = 0,15{\text{ lít = 150 ml}}\)
\( \to {V_{dd{\text{ HCl}}}} = 150.1,1 = 165{\text{ gam}}\)
BTKL:
\({m_{MgC{O_3}}} + {m_{CuC{O_3}}} + {m_{dd{\text{ HCl}}}} = {m_{dd}} + {m_{C{O_2}}}\)
\( \to {m_{dd}} = 14,6 + 165 - 0,15.44 = 173{\text{ gam}}\)
\({n_{MgC{l_2}}} = {n_{MgC{O_3}}} = 0,1{\text{ mol}}\)
\({n_{CuC{l_2}}} = {n_{CuC{O_3}}} = 0,05{\text{ mol}}\)
\( \to {m_{MgC{l_2}}} = 0,1.(24 + 35,5.2) = 9,5{\text{ gam}}\)
\({m_{CuC{l_2}}} = 0,05.(64 + 35,5.2) = 6,75{\text{ gam}}\)
\( \to C{\% _{MgC{l_2}}} = \frac{{9,5}}{{173}} = 5,49\% \)
\(C{\% _{CuC{l_2}}} = \frac{{6,75}}{{173}} = 3,9\% \)
