Đáp án:
\( \% {m_{Mg}} = 54,54\% ; \% {m_{MgO}} = 45,46\% \)
Giải thích các bước giải:
Gọi số mol \(Mg;MgO\) lần lượt là \(x;y\)
\( \to 24x + 40y = 8,8{\text{ gam}}\)
Phản ứng xảy ra:
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
\(MgO + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}O\)
Ta có:
\({n_{{H_2}}} = {n_{Mg}} = x = \frac{{4,48}}{{22,4}} = 0,2 \to y = 0,1\)
\( \to {m_{Mg}} = 0,2.24 = 4,8{\text{ gam}}\)
\( \to \% {m_{Mg}} = \frac{{4,8}}{{8,8}} = 54,54\% \to \% {m_{MgO}} = 45,46\% \)
\({n_{HCl}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,2.2 + 0,1.2 = 0,6{\text{ mol}}\)
\( \to {m_{HCl}} = 0,6.36,5 = 21,9{\text{ gam}}\)
\( \to {m_{dd{\text{ HCl}}}} = \frac{{21,9}}{{7,3\% }} = 300{\text{ gam}}\)
BTKL:
\({m_{hh}} + {m_{dd{\text{ HCl}}}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 8,8 + 300 = {m_{dd}} + 0,2.2 \to {m_{dd}} = 308,4{\text{ gam}}\)
\({n_{MgC{l_2}}} = {n_{Mg}} + {n_{MgO}} = 0,2 + 0,1 = 0,3{\text{ mol}}\)
\( \to {m_{MgC{l_2}}} = 0,3.(24 + 35,5.2) = 28,5{\text{ gam}}\)
\( \to C{\% _{MgC{l_2}}} = \frac{{28,5}}{{308,4}} = 9,24\% \)
