1)
Gọi số mol \(CuO;Fe_2O_3\) trong hỗn hợp lần lượt là \(x;y\)
\(CuO + 2HCl\xrightarrow{{}}CuC{l_2} + {H_2}O\)
\(F{e_2}{O_3} + 6HCl\xrightarrow{{}}2FeC{l_3} + 3{H_2}O\)
Ta có:
\({n_{CuC{l_2}}} = {n_{CuO}} = x{\text{ mol}}\)
\({n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 2y{\text{ mol}}\)
Vì số mol 2 muối bằng nhau.
\( \to {n_{CuC{l_2}}} = {n_{FeC{l_3}}} \to x = 2y\)
Giải được:
\(x=0,25;y=0,125\)
\( \to {m_{muối}} = {m_{CuC{l_2}}} + {m_{FeC{l_3}}}\)
\( = 0,25.(64 + 35,5.2) + 0,25.(56 + 35,5.3) = 74,375{\text{ gam}}\)
2)
Phản ứng xảy ra:
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
Ta có:
\({n_{Mg}} = \frac{{9,6}}{{24}} = 0,4{\text{ mol = }}{{\text{n}}_{MgC{l_2}}} = {n_{{H_2}}}\)
\( \to {m_{MgC{l_2}}} = 0,4.(24 + 35,5.2) = 38{\text{ gam}}\)
BTKL:
\({m_{Mg}} + {m_{dd{\text{ HCl}}}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 9,6 + 120 = {m_{dd}} + 0,4.2 \to {m_{dd}} = 128,8{\text{ gam}}\)
\( \to C{\% _{MgC{l_2}}} = \frac{{38}}{{128,8}} = 29,5\% \)
