36)
Cho hỗn hợp tác dụng với \(HCl\) thì chỉ có \(Zn\) phản ứng
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({m_{{H_2}}} = 0,1{\text{ gam}} \to {{\text{n}}_{{H_2}}} = \frac{{0,1}}{2} = 0,05{\text{ mol}} = {n_{Zn}}\)
\( \to {m_{Zn}} = 0,05.65 = 3,25{\text{ gam}}\)
\( \to \% {m_{Zn}} = \frac{{3,25}}{5} = 65\% \to \% {m_{Cu}} = 35\% \)
\({n_{HCl}} = 2{n_{{H_2}}} = 0,05.2 = 0,1{\text{ mol}}\)
\( \to {C_{M{\text{ HCl}}}} = \frac{{0,1}}{2} = 0,05M\)
\( \to {n_{ZnC{l_2}}} = {n_{Zn}} = 0,05{\text{ mol}}\)
\( \to {C_{M{\text{ ZnC}}{{\text{l}}_2}}} = \frac{{0,05}}{2} = 0,025M\)
37)
Cho hỗn hợp tác dụng với \(HCl\)
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
\(FeO + 2HC\xrightarrow{{}}FeC{l_2} + {H_2}O\)
Ta có:
\({n_{{H_2}}} = \frac{{1,12}}{{22,4}} = 0,05{\text{ mol = }}{{\text{n}}_{Fe}}\)
\( \to {m_{Fe}} = 0,05.56 = 2,8{\text{ gam}} \to {{\text{m}}_{FeO}} = 10 - 2,8 = 7,2{\text{ gam}}\)
\(\% {m_{Fe}} = \frac{{2,8}}{{10}} = 28\% \to \% {m_{FeO}} = 72\% \)
Ta có:
\({n_{FeO}} = \frac{{7,2}}{{56 + 16}} = 0,1{\text{ mol}}\)
\( \to {n_{HCl}} = 2{n_{Fe}} + 2{n_{FeO}} = 0,05.2 + 0,1.2 = 0,3{\text{ mol}}\)
\( \to {m_{HCl}} = 0,3.36,5 = 10,95{\text{ gam}}\)
\( \to {m_{dd\;{\text{HCl}}}} = \frac{{10,95}}{{30\% }} = 36,5{\text{ gam}}\)
3)
Cho hỗn hợp tác dụng với \(HCl\) thì xảy ra các phản ứng:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
\(F{e_2}{O_3} + 6HCl\xrightarrow{{}}2FeC{l_3} + 3{H_2}O\)
Gọi số mol \(Fe;Fe_2O_3\) lần lượt là \(x;y\)
\( \to {m_{hh}} = 56x + 160y = 13,6\)
Ta có:
\({n_{HCl}} = 0,5.1 = 0,5{\text{ mol = 2}}{{\text{n}}_{Fe}} + 6{n_{F{e_2}{O_3}}} = 2x + 6y\)
Giải được: \(x=0,1;y=0,05\)
\( \to {m_{Fe}} = 0,1.56 = 5,6{\text{ gam}}\)
\( \to \% {m_{Fe}} = \frac{{5,6}}{{13,6}} = 41,17\% \to \% {m_{F{e_2}{O_3}}} = 58,83\% \)
Ta có:
\({n_{{H_2}}} = {n_{Fe}} = 0,1{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
Ta có:
\({n_{FeC{l_2}}} = {n_{Fe}} = 0,1{\text{ mol;}}{{\text{n}}_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,1{\text{ mol}}\)
\( \to {m_{muối}} = {m_{FeC{l_2}}} + {m_{FeC{l_3}}}\)
\( = 0,1.(56 + 35,5.2) + 0,1.(56 + 35,5.3) = 28,95{\text{ gam}}\)
