a, 2K + 2H2O--> 2KOH + H2

B là Cu , Fe . Khí C = H2 A: ddKOH

Fe + 2HCl--> FeCl2 + H2

TA có nH2=22,4/22,4=1

=> nK=1/2=0,5

Ta có cran 6,6 g còn lại là Cu

=> nHCl= 0,5.0,4=0,2

=> nFe= 0,2/2=0,1

=> mhh=0,1.56 + 0,5.39 + 6,6=31,7 g

=> %mK= 0,5.39.100/31,7=61,5%

%mFe= 0,1.56.100/31,7=17,66 %

%mCu= 6,6.100/31,7=20,82%

b, FexOy + yH2--t0-> xFe + yH2O

Ta có nH2=1

=> 1/y=nFexOy

<=> 1/y=5,8/(56x + 16y)

<=> 56x + 16y= 5,8y

<=> 56x = -10,2y ????

a)

\(n_{H_2\left(1\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂ (1)

             0,6<----------------------0,3

=> m_Na = 0,6.23 = 13,8 (g)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,1<-0,2

=> m_Fe = 0,1.56 = 5,6 (g)

m_Cu = 10 (g)

\(\left\{{}\begin{matrix}\%Na=\dfrac{13,8}{13,8+5,6+10}.100\%=46,94\%\\\%Fe=\dfrac{5,6}{13,8+5,6+10}.100\%=19,05\%\\\%Cu=\dfrac{10}{13,8+5,6+10}.100\%=34,01\%\end{matrix}\right.\)

b)

PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

             \(\dfrac{0,3}{y}\)<--0,3

=> \(M_{Fe_xO_y}=\dfrac{17,4}{\dfrac{0,3}{y}}=58y\left(g/mol\right)\)

=> 56x = 42y

=> \(\dfrac{x}{y}=\dfrac{3}{4}\) => CTHH: Fe₃O₄

a)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

            0,6<----------------------0,3

             Fe + 2HCl --> FeCl₂ + H₂

            0,1<--0,2

=> \(\left\{{}\begin{matrix}m_{Na}=0,6.23=13,8\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Cu}=10\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{13,8}{13,8+5,6+10}.100\%=46,94\%\\\%m_{Fe}=\dfrac{5,6}{13,8+5,6+10}.100\%=19,05\%\\\%m_{Cu}=\dfrac{10}{13,8+5,6+10}.100\%=34,01\%\end{matrix}\right.\)

b)
PTHH: Fe_xO_y + yH₂ --t^o--> xFe + yH₂O

            \(\dfrac{0,3}{y}\)<--0,3

=> \(M_{Fe_xO_y}=56x+16y=\dfrac{17,4}{\dfrac{0,3}{y}}\left(g/mol\right)\)

=> \(\dfrac{x}{y}=\dfrac{3}{4}\)

=> CTHH: Fe₃O₄

Dung dịch A:  dd KOH

Rắn B: Cu, Fe

Khí C: H₂

Các PTHH:

\(2K+2H_2O\rightarrow2KOH+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{HCl}=\dfrac{58,4.15\%}{36,5}=0,24\left(mol\right)\\ Fe+2HCl\rightarrow\left(t^o\right)FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{0,24}{2}=0,12\left(mol\right)\\ \Rightarrow V1=V_{H_2\left(đktc\right)}=0,12.22,4=2,688\left(l\right)\\ x=m_{Cu}=m_{hhA}-m_{Fe}=15,68-0,12.56=8,96\left(g\right)\\ b,n_{Cu}=\dfrac{8,96}{64}=0,14\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ Cu+Cl_2\rightarrow\left(t^o\right)CuCl_2\\ n_{Cl_2}=\dfrac{3}{2}.n_{Fe}+n_{Cu}=\dfrac{3}{2}.0,12+0,14=0,32\left(mol\right)\\ \Rightarrow V2=V_{Cl_2\left(đktc\right)}=0,32.22,4=7,168\left(l\right)\\ y=m_{muối}=m_{AlCl_3}+m_{CuCl_2}=0,12.133,5+0,14.135=34,92\left(g\right)\)

nB=0,05.
B=3,2/0,05=64.(Cu).
nA=0,05.
A=3,25/0,05=65.(Zn).
F:Cu(N03)2.
mF=9,4g.
mKHI=3,24.
Goj n02=x.
=>216x=3,24.
x=0,015.
V=1,68l.
Zn - Cu.
Klg gjam 0,1g.
=>nCu(N03)=0,1.
CM=0,1/0,4=0,25M.

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