a)

\(2K+2H_2O\rightarrow2KOH+H_2\)

\(2K+2HCl\rightarrow2KCl+H_2\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(4K+O_2\underrightarrow{t^o}2K_2O\)

\(2Mg+O_2\underrightarrow{t^o}2MgO\)

b)

P1: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2K + 2H₂O --> 2KOH + H₂

          0,3<--------------------0,15

=> n_K(X) = 0,3.3 = 0,9 (mol)

P2:

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: 2K + 2HCl --> 2KCl + H₂

           0,3------------------->0,15

             Mg + 2HCl --> MgCl₂ + H₂

           0,25<-------------------0,25

=> n_Mg(X) = 0,25.3 = 0,75 (mol)

P3:

PTHH: 4K + O₂ --t^o--> 2K₂O

          0,3------------->0,15

            2Mg + O₂ --t^o--> 2MgO

          0,25--------------->0,25

=> m_Ag(P3) = 34,9 - 0,15.94 - 0,25.40 = 10,8 (g)

=> m_Ag(X) = 10,8.3 = 32,4 (g)

a = m_X = 0,9.39 + 0,75.24 + 32,4 = 85,5 (g)

 a) PTHH: $Fe+2HCl\to FeC{l}_2+H_2\uparrow$

b) Ta có: $n_{H_2}=\frac{}{}=0,25\left(mol\right)=n_{FeC{l}_2}$

$\Rightarrow m_{FeC{l}_2}=0,25\cdot 127=31,75\left(g\right)$

c) Theo PTHH: $n_{H_2}=n_{Fe}=0,25mol$ $\Rightarrow m_{Fe}=0,25\cdot 56=14\left(g\right)$

a) PTHH: $Fe+2HCl\to FeC{l}_2+H_2\uparrow$

b) Ta có: $n_{H_2}=\frac{}{}=0,25\left(mol\right)=n_{FeC{l}_2}$ 

⇒mFeCl2=0,25⋅127=31,75(g)

c) Theo PTHH: $n_{H_2}=n_{Fe}=0,25mol$ $\Rightarrow m_{Fe}=0,25\cdot 56=14\left(g\right)$

$\Rightarrow \%m_F$

- Đăt nAl = x mol ; nFe = y mol

=> 27x + 56y = 16,8 (I)

nH2 = 0,5 mol

- PTHH: 2Al (x) + 6HCl ----> 2AlCl3 + 3H2 (1,5x) (1)

Fe (y) + 2HCl -----> FeCl2 + H2 (y) (2)

- Theo PTHH: nH2 = 1,5x + y = 0,5 (II)

- Giải hệ PT (I;II) => \(\left\{{}\begin{matrix}x=\dfrac{56}{285}\left(mol\right)\\y=\dfrac{39}{190}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}mAl=\dfrac{504}{95}\left(gam\right)\\mFe=\dfrac{1092}{95}\left(gam\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%mAl=31,6\%\\\%mFe=68,4\%\end{matrix}\right.\)

b) - Bảo toàn H: => nHCl = 1 mol

=> V HCl = 2 lít

c) - Bảo toàn Al: => nAlCl3 = \(\dfrac{56}{285}\left(mol\right)\)

- Bảo toàn Fe: => nFeCl2 = \(\dfrac{39}{190}\left(mol\right)\)

=> m muối sau pư = \(\dfrac{56}{285}.133,5+\dfrac{39}{190}.127=52,3\left(gam\right)\)

V HCl = 1/2 = 0,5 lít ( lộn chỗ đó)

mai mình thi r giúp mình vs

\(a.2Al+6HCl->2AlCl_3+3H_2\)

        0,1.....................................0,1..........0,15

b. n Al = \(\frac{2,7}{27}=0,1mol\)

m AlCl3 = 0,1(27+35,5.3)=13,35 g

V H2 = 0,15.22,4=3,36 lít

Fe+2HCl\(\rightarrow\)FeCl₂+H₂

\(n_{Fe}=\dfrac{14}{56}=0,25mol\)

\(n_{H_2}=n_{Fe}=0,25mol\)

\(n_{FeCl_2}=n_{Fe}=0,25mol\)

\(m_{FeCl_2}=0,25.127=31,75gam\)

H₂+CuO\(\overset{t^0}{\rightarrow}\)Cu+H₂O

\(n_{CuO}=\dfrac{25,6}{80}=0,32mol\)

Tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,32}{1}\)\(\rightarrow\)CuO dư

\(n_{Cu}=n_{CuO}=n_{H_2}=0,25mol\)

\(m_{Cu}=0,25.64=16gam\)

\(n_{CuO\left(dư\right)}=0,32-0,25=0,07mol\)

\(m_{CuO}=0,07.80=5,6gam\)

giúp minh với các bạn 

a, Cu không tác dụng với dd HCl.

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)

\(\Rightarrow m_{Cu}=19,4-13=6,4\left(g\right)\)

c, Ta có: \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{19,4}.100\%\approx67,01\%\\\%m_{Cu}\approx32,99\%\end{matrix}\right.\)

Bạn tham khảo nhé!