Đáp án:
\(C{\% _{AlC{l_3}}} = 1,38\% \)
\(C{\% _{FeC{l_2}}} = 0,658\% \)
Giải thích các bước giải:
Gọi số mol \(Al;Fe\) lần lượt là \(x;y\)
\( \to 27x + 56y = 11\)
Cho hỗn hợp tác dụng với \(HCl\)
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{8,96}}{{22,4}} = 0,4{\text{ mol = }}\frac{3}{2}{n_{Al}} + {n_{Fe}} = 1,5x + y\)
Giải được: \(x=0,2;y=0,1\)
\({n_{HCl}} = 2{n_{{H_2}}} = 0,8{\text{ mol}}\)
\( \to {V_{dd{\text{HCl}}}} = \frac{{0,8}}{{0,5}} = 1,6{\text{ lít = 1600 ml}}\)
\( \to {m_{dd\;{\text{HCl}}}} = 1600.1,2 = 1920{\text{ gam}}\)
BTKL:
\({m_{kl}} + {m_{dd{\text{ HCl}}}} = {m_{dd}} + {m_{{H_2}}}\)
\( \to 11 + 1920 = {m_{dd}} + 0,4.2\)
\( \to {m_{dd}} = 1930,2{\text{ gam}}\)
\({n_{AlC{l_3}}} = {n_{Al}} = 0,2{\text{ mol;}}{{\text{n}}_{FeC{l_2}}} = {n_{Fe}} = 0,1{\text{ mol}}\)
\( \to {m_{AlC{l_3}}} = 0,2.(27 + 35,5.3) = 26,7{\text{ gam}}\)
\({m_{FeC{l_2}}} = 0,1.(56 + 35,5.2) = 12,7{\text{ gam}}\)
\( \to C{\% _{AlC{l_3}}} = \frac{{26,7}}{{1930,2}} = 1,38\% \)
\(C{\% _{FeC{l_2}}} = \frac{{12,7}}{{1930,2}} = 0,658\% \)
