Gọi n Na = x ; ⁿBa= y trong 0,297g hợp kim

=> 23x + 137y = 0,297    (I)

2Na + 2H2O -------> 2NaOH + H2      (1)

   x  ---------->                 x    --->  0,5 x               (mol)

Ba + 2H2O --------> Ba(OH)2 + H2     (2)

  y ----------------->      y       -------> y                   (mol)

dd X gồm NaOH và Ba(OH)₂ 

Khí Y là H2

NaOH + HCl -------> NaCl + H2O     (3)

    x   --->   x  ------->        x                           (mol)

Ba(OH)2 + 2HCl -------> BaCl2 +2 H2O     (4)

       y -----> 2y  -------->          y                                (mol)

m muối = m NaCl + m BaCl2

==> 58,5x + 208y = 0,4745   (ll)

Từ (l) và (ll) => x= 0,001 và y = 0,002

Theo pt (1)(2) ⁿH₂= 0,5x + y = 0,0025 (mol)

=> V H2 = 0,0025 . 22,4 = 0,056 (l)

Theo pt(3)(4) n HCl = x + 2y =0,005  (mol)

=> C_M(HCl)=\(\frac{0,005}{0,05}\)=0,1 (M)

%^mNa = \(\frac{0,001.23}{0,297}\). 100% = 7,74%

%^mBa = 100% - 7,74% = 92,26%

a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

Gọi \(\left\{{}\begin{matrix}n_{Al}:x\left(mol\right)\\n_{Fe}:y\left(mol\right)\end{matrix}\right.\)

Ta có : \(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\)

=> x=0,2 ; y=0,1

 \(\%m_{Al}=\dfrac{0,2.27}{11}.100==49,09\%\)

\(\%m_{Fe}=50,91\%\)

b) \(\Sigma n_{HCl}=3x+2y=0,8\left(mol\right)\)

=> \(V_{HCl}=\dfrac{0,8}{2}=0,4\left(lít\right)\)

c) \(CM_{AlCl_3}=\dfrac{0,2}{0,4}=0,5M\)

\(CM_{FeCl_2}=\dfrac{0,1}{0,4}=0,25M\)

\(\text{Gọi số mol của Na là x(mol)},\text{số mol của Ba là y(mol)}\)

\(\text{=> 23x +137y=0.297(1)}\)

PTPU với Hiđrô

2Na +2H2O -----> 2NaOH+ H2

x.................................. x..........x/2

Ba + 2H2O -------> Ba(OH)2 +2H2O

y.....................................y................2y

PTPU với HCl:

NaOH + HCl ------> NaCl +H2O

x................x.......................x (mol)

Ba(OH)2 +2HCl ----> BaCl2+2H2O

y.....................2y....................y (mol)

\(\text{m muối = 58,5x+208y=0,4745 (2) }\)

Giải hệ (1) và (2) => x=0,001 ; y=0,002

\(\text{V khí thoát ra= (0,001/2+0,002).22,4=0,056(l)}\)

\(\text{nHCl=0,001+0,002.2=0,005(mol)}\)

\(\text{CM HCl= 0,005/50=0,0001M}\)

\(\left\{{}\begin{matrix}\text{mNa=23.0,001=0,023(g)}\\\text{mBa =137.0,002=0,274(g)}\end{matrix}\right.\)

\(n_{Zn}=a\left(mol\right),n_{Al}=b\left(mol\right)\)

\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+H_2\)

\(n_{H_2}=a+1.5b=0.4\left(mol\right)\left(1\right)\)

\(m_{Muối}=m_{ZnCl_2}+m_{AlCl_3}=136a+133.5b=40.3\left(g\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.1,b=0.2\)

\(m_{hh}=0.1\cdot65+0.2\cdot27=11.9\left(g\right)\)

\(\%Zn=\dfrac{0.1\cdot65}{11.9}\cdot100\%=54.62\%\)

\(\%Al=100-54.62=45.38\%\)

\(n_{Zn}=\dfrac{4,55}{65}=0,07(mol)\\ Zn+2HCl\to ZnCl_2+H_2\\ a,n_{HCl}=0,14(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,14}{0,2}=0,7M\\ b,n_{H_2}=0,07(mol)\\ \Rightarrow V_{H_2}=0,07.22,4=1,568(l)\\ c,n_{ZnCl_2}=0,07(mol)\\ \Rightarrow m_{ZnCl_2}=0,07.136=9,52(g)\\ c,ZnCl_2+2AgNO_3\to 2AgCl\downarrow+Zn(NO_3)_2\)

\(m_{dd_{ZnCl_2}}=200.0,8+4,55-0,07.2=164,41(g)\\ n_{AgCl}=0,14(mol);n_{Zn(NO_3)_2}=0,07(mol)\\ \Rightarrow C\%_{Zn(NO_3)_2}=\dfrac{0,07.189}{164,41+200-0,14.143,5}.100\%=3,84%\)

n_HCl=5.10^-3 mol

2Na + 2H₂O --> 2NaOH + H₂

_{x mol                     x mol        1/2 mol}

Ba + 2H₂O --> Ba(OH)₂ + H₂

_{y mol                   y mol          y mol}

NaOH + HCl --> NaCl + H₂O

x mol      x mol

Ba(OH)₂ + 2HCl--> BaCl₂ + H₂O

y mol           2y mol

Ta duoc: 23x + 137y =0,297   (1)

                 x + 2y =5.10^-3    (2)

Tu (1) va (2) ta duoc => x= 10^-3

                                   => y= 2.10^-3

a/ m_Na= 10^-3.23=0,023g

     m_Ba=2.10^-3.137=0,274g

b/ nH₂= 10^-6 mol

H_{2           +}          O₂   -->          H₂O

10^-6 mol          10^-6 mol

V_O2= 10^-6. 22,4=2,24.10^-5  lit

V_KK= 2,24.10^-5.100/20=1,12.10^-4 lit