1)

Gọi số mol \(Zn; Cu\) lần lượt là \(x;y\).

\( \to 65x + 64y = 12,9\)

Cho hỗn hợp tác dụng với axit loãng

\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)

Ta có:

\({n_{{H_2}S{O_4}}} = 0,1.1 = 0,1{\text{ mol = }}{{\text{n}}_{Zn}}\) 

\( \to {m_{Zn}} = 0,1.65 = 6,5{\text{ gam}} \to {{\text{m}}_{Cu}} = 6,4{\text{ gam}}\)

\( \to \% {m_{Zn}} = \frac{{6,5}}{{12,9}} = 50,39\%  \to \% {m_{Cu}} = 49,61\% \)

Ta có: 

\({n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,1{\text{ mol}}\)

\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

\({n_{ZnS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,1\)

\( \to {C_{M{\text{ ZnS}}{{\text{O}}_4}}} = \frac{{0,1}}{{0,1}} = 1M\)

2)

Phản ứng xảy ra:

\(F{e_2}{O_3} + 3{H_2}\xrightarrow{{{t^o}}}2Fe + 3{H_2}O\)

Rắn \(A\) chứa \(Fe\) và \(CaO\).

\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}} \to {{\text{n}}_{F{e_2}{O_3}}} = \frac{{{n_{{H_2}}}}}{3} = \frac{{0,1}}{3}\)

\( \to {m_{F{e_2}{O_3}}} = \frac{{0,1}}{3}.160 = \frac{{16}}{3}\)

\( \to \% {m_{F{e_2}{O_3}}} = \frac{{\frac{{16}}{3}}}{{21,6}} = 24,69\%  \to \% {m_{CaO}} = 75,31\% \)

Giải thích các bước giải:

Câu 1:

\(\begin{array}{l} a,\ PTHH:Zn+H_2SO_4\to ZnSO_4+H_2↑\\ b,\ n_{H_2SO_4}=0,1\times 1=0,1\ mol.\\ Theo\ pt:\ n_{Zn}=n_{H_2SO_4}=0,1\ mol.\\ ⇒\%m_{Zn}\dfrac{0,1\times 65}{12,9}=50,39\%\\ \%m_{Cu}=100\%-50,39\%=49,61\%\\ c,\ Theo\ pt:\ n_{ZnSO_4}=n_{H_2SO_4}=0,1\ mol.\\ \Rightarrow C_{M_{ZnSO_4}}=\dfrac{0,1}{0,1}=1\ M.\end{array}\)

Câu 2:

\(\begin{array}{l} a,\ PTHH:Fe_2O_3+3H_2\xrightarrow{t^o} 2Fe+3H_2O\\ \text{Các chất trong A gồm Fe, CaO.}\\ b,\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\ mol.\\ Theo\ pt:\ n_{Fe_2O_3}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{30}\ (mol).\\ ⇒\%m_{Fe_2O_3}=\dfrac{\dfrac{1}{30}\times 160}{21,6}\times 100\%=24,69\%\\ \%m_{CaO}=100\%-24,69\%=75,31\%\end{array}\)

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