Hóa học

B1:

Đặt nNa = x mol; nAl = y mol

Thí nghiệm 1:

Na + H2O → NaOH + ½ H2

x →                  x        x/2 mol       

NaOH + Al + H2O → NaAlO2+ 3/2 H2

x →                                        3/2x mol

Ta có: nH2 = x/2 + 3x/2 = 2x mol → VH2 = V1

Thí nghiệm 2:

Na + H2O → NaOH + ½ H2

x →                 x        x/2 mol       

NaOH + Al + H2O → NaAlO2+ 3/2 H2

               y →                          3/2y mol

Ta có: nH2 = x/2 + 3/2y → VH2 = V2

+ Nếu x = y thì V1 = V2

+ Nếu x < y thì V2 > V1

Vậy V1 ≤ V2

Làm nốt bài 2 còn lại bạn nhé!

\(4NH_3+7O_2\xrightarrow[Pt]{t^o}4NO+6H_2O\)

V------>1,75V---->V

\(2NO+O_2\rightarrow2NO_2\)

V----->0,5V--->V

\(4NO_2+2H_2O+O_2\rightarrow4HNO_3\)

V----------------->0,25V

\(\rightarrow V'=1,75V+0,5V+0,25V=\dfrac{5}{2}V\\ \Leftrightarrow\dfrac{V'}{V}=\dfrac{5}{2}\)

\(n_{Al\left(OH\right)_3}=\dfrac{7,8}{78}=0,1\left(mol\right)\)

- TH₁: Nếu kết tủa bị tan một phần

PTHH:

\(HCl+NaOH\rightarrow NaCl+H_2O\)

a------->a

\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\)

0,2----->0,6

\(n_{Al\left(OH\right)_3\left(tan\right)}=0,2-0,1=0,1\left(mol\right)\)

\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)

0,1--------->0,1

=> 0,6 + 0,1 + a = 0,8

=> a = 0,1 

- TH₂: Kết tủa không bị tan

\(HCl+NaOH\rightarrow NaCl+H_2O\)

a------->a

\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\)

               0,3<--------0,1

=> a + 0,3 = 0,8

=> a = 0,5

Gọi số mol Na₂O, Al₂O_3(bđ) là a, b (mol)

\(n_{Al_2O_3\left(pư\right)}=b-\dfrac{10,2}{102}=b-0,1\left(mol\right)\)

PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

                 a---------------->2a

            \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)

                 a<--------2a---------->2a

\(\left\{{}\begin{matrix}n_{Al_2O_3\left(pư\right)}=a=b-0,1\\m_{NaAlO_2}=164a\left(g\right)\end{matrix}\right.\)

m_{dd sau pư} = 62a + 102a + 200 

= 164a + 200

Ta có: \(C\%_{NaAlO_2}=\dfrac{164a}{164a+200}.100\%=10\%\)

=> a = \(\dfrac{50}{369}\left(mol\right)\Rightarrow b=\dfrac{869}{3690}\left(mol\right)\)

\(\left\{{}\begin{matrix}\%m_{Na_2O}=\dfrac{\dfrac{50}{369}.62}{\dfrac{50}{369}.62+\dfrac{869}{3690}.102}.100\%=25,91\%\\\%m_{Al_2O_3}=\dfrac{\dfrac{869}{3690}.102}{\dfrac{869}{3690}.102+\dfrac{50}{369}.62}.100\%=74,09\%\end{matrix}\right.\)

\(Na_2SO_3+2HCl\rightarrow2NaCl+SO_2+H_2O\)

0,4               0,8          0,8           0,4     0,8   (mol)

\(n_{SO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

\(m_{HCl}=0,8.36,5=29,2\left(g\right)\)

=> mddHCl = ( 29,2.100 ) / 21,9% = 133\(\dfrac{1}{3}\left(g\right)\)

\(m_{Na_2SO_3}=0,8.126=100,8\left(g\right)\)

ta cho hỗn hợp vào NaOH đặc sau đó vớt kết tủa thì thu đc Mg(OH)2

Al(OH)3+NaOH->NaAlO2+2H2O

Zn(OH)2+2NaOH->Na2ZnO2+2H2O

M₂O_x + xH₂SO₄ ->M₂(SO₄)x + xH₂O

1………………x………………..1

mdđ muối = mM2Ox + mddH2SO4 = 2M + 16x + \(\dfrac{98x}{20\%}\) = 2M + 506x

C%M₂(SO4)_x = \(\dfrac{2M+96x}{2M+506x}\)= 21,75%

=> x = 3, M = 27: Oxit là Al2O3

a) Gọi số mol Al, Fe là a, b (mol)

=> 27a + 56b = 11 (1)

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

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

             a------>3a------->a----->1,5a

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

             b----->2b------->b------>b

=> 1,5a + b = 0,4 (2)

(1)(2) => a = 0,2 (mol); b = 0,1 (mol)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{11}.100\%=49,1\%\\\%m_{Fe}=\dfrac{0,1.56}{11}.100\%=50,9\%\end{matrix}\right.\)

n_HCl = 3a + 2b = 0,8 (mol)

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

\(\left\{{}\begin{matrix}C_{M\left(AlCl_3\right)}=\dfrac{0,2}{0,4}=0,5M\\C_{M\left(FeCl_2\right)}=\dfrac{0,1}{0,4}=0,25M\end{matrix}\right.\)

b)

5,5 (g) hh chứa \(\left\{{}\begin{matrix}Al:0,1\left(mol\right)\\Fe:0,05\left(mol\right)\end{matrix}\right.\)

PTHH: \(2Al+6H_2SO_{4\left(đ,n\right)}\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)

             0,1----------------------->0,05----->0,15

            \(2Fe+6H_2SO_{4\left(đ,n\right)}\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

            0,05-------------------->0,025----->0,075

=> V = (0,15 + 0,075).22,4 = 5,04 (l)

=> a = 0,05.342 + 0,025.400 = 27,1 (g)

Gọi n Al = a ( mol ) , n Fe = b ( mol )

Có: n H₂ = 0,4 ( mol )

   PTHH

 2AL + 6HCL ===> 2ALCL₃ + 3H₂

   a--------------------------------------a

 Fe + 2HCl ====> FeCl₂ + H2

  b------------------------------------b

Ta có hpt:

\(\left\{{}\begin{matrix}27a+56b=111,5\\a+b=0,4\end{matrix}\right.\)

=>a=0,2mol , b=0,1 mol

=> m Al = 5,4 ( g ) ; m Fe = 5,6 ( g )

=>%Al=49%

=>%Fe=51%

CM _AlCl3=\(\dfrac{0,2}{0,2}\)=1M

CM _FeCl2=\(\dfrac{0,1}{0,2}\)=0,5M

C₂H₅NO₂+O₂-to>2CO₂+H₂O+0,5N₂

0,33----------------------------------------0,167

n _C2H5NO2=0,33 mol

=>VN2=0,167.22,4=3,73l

a)

Gọi số mol Mg, Al, Zn là a, b, c (mol)

=> 24a + 27b + 65c = 35 (1)

\(n_{H_2}=\dfrac{19,04}{22,4}=0,85\left(mol\right)\)

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

              a----------------->a----->a

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

               b------------------>b---->1,5b

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

               c------------------>c----->c

=> \(\left\{{}\begin{matrix}a+1,5b+c=0,85\left(2\right)\\1,5b=2a\left(3\right)\end{matrix}\right.\)

(1)(2)(3) => a = 0,15 (mol); b = 0,2 (mol); c = 0,4 (mol)

\(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,15.24}{35}.100\%=10,286\%\\\%m_{Al}=\dfrac{0,2.27}{35}.100\%=15,428\%\\\%m_{Zn}=\dfrac{0,4.65}{35}.100\%=74,286\%\end{matrix}\right.\)

b)

A chứa \(\left\{{}\begin{matrix}MgCl_2:0,15\left(mol\right)\\AlCl_3:0,2\left(mol\right)\\ZnCl_2:0,4\left(mol\right)\\HCl\left(dư\right)\end{matrix}\right.\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

\(2NaOH+MgCl_2\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)

\(3NaOH+AlCl_3\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\)

\(2NaOH+ZnCl_2\rightarrow Zn\left(OH\right)_2\downarrow+2NaCl\)

\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)

\(Zn\left(OH\right)_2+2NaOH\rightarrow Na_2ZnO_2+2H_2O\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

Vậy chất rắn B là MgO

Theo PTHH: \(n_{MgO}=0,15\left(mol\right)\Rightarrow m_{MgO}=0,15.40=6\left(g\right)\)

- Xét TN1:

\(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

          0,04<--------------------0,06

- Xét TN2:

\(n_{SO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: \(Cu+2H_2SO_{4\left(đ\right)}\rightarrow CuSO_4+SO_2+2H_2O\)

            0,4<----------------------------0,4

=> Trong m (g) hỗn hợp chứa \(\left\{{}\begin{matrix}Al:0,04\left(mol\right)\\Cu:0,2\left(mol\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,04.27}{0,04.27+0,2.64}.100\%=7,78\%\\\%m_{Cu}=\dfrac{0,2.64}{0,04.27+0,2.64}.100\%=92,22\%\end{matrix}\right.\)