Đáp án:
a) \(\% {m_{Fe}} = 29,58\% ;\% {m_{CuO}} = 70,42\% \)
b) \({{\text{C}}_{M{\text{ HCl}}}} = 4M\)
c) \({m_{kt}} = 19{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
\(CuO + 2HCl\xrightarrow{{}}CuC{l_2} + {H_2}O\)
Ta có:
\({n_{Fe}} = {n_{{H_2}}} = \frac{{1,68}}{{22,4}} = 0,075{\text{ mol}} \to {{\text{m}}_{Fe}} = 0,075.56 = 4,2{\text{ gam}} \to {{\text{m}}_{CuO}} = 14,2 - 4,2 = 10{\text{ gam}} \to {{\text{n}}_{CuO}} = \frac{{10}}{{64 + 16}} = 0,125{\text{ mol}}\)
\(\to \% {m_{Fe}} = \frac{{4,2}}{{14,2}} = 29,58\% \to \% {m_{CuO}} = 70,42\% \)
Ta có: \({n_{HCl}} = 2{n_{Fe}} + 2{n_{CuO}} = 0,075.2 + 0,125.2 = 0,4{\text{ mol}} \to {{\text{C}}_{M{\text{ HCl}}}} = \frac{{0,4}}{{0,1}} = 4M\)
B chứa \(FeC{l_2}\) 0,075 mol và \(CuC{l_2}\) 0,125 mol. Cho B tác dụng với NaOH dư.
\(FeC{l_2} + 2NaOH\xrightarrow{{}}Fe{(OH)_2} + 2NaCl\)
\(CuC{l_2} + 2NaOH\xrightarrow{{}}Cu{(OH)_2} + 2NaCl\)
\(\to {n_{Fe{{(OH)}_2}}} = {n_{FeC{l_2}}} = 0,075{\text{ mol; }}{{\text{n}}_{Cu{{(OH)}_2}}} = {n_{CuC{l_2}}} = 0,125{\text{ mol}}\)
\(\to {m_{kt}} = 0,075.(56 + 17.2) + 0,125.(64 + 17.2) = 19{\text{ gam}}\)
