\(a.n_{HCl}=0,1.1=0,1\left(mol\right)\\ n_{H_2SO_4}=0,5.0,1=0,05\left(mol\right)\\ \left[HCl\right]=\dfrac{0,1}{0,1+0,1}=0,5\left(M\right)\\ \left[H_2SO_4\right]=\dfrac{0,05}{0,1+0,1}=0,25\left(M\right)\\ \left[H^+\right]=0,5+0,25.2=1\left(M\right)\\ \left[SO^{2-}_4\right]=\left[H_2SO_4\right]=0,25\left(M\right)\\ \left[Cl^-\right]=\left[HCl\right]=0,5\left(M\right)\)

\(b.BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ n_{BaSO_4}=n_{H_2SO_4}=0,05\left(mol\right)\\ m_{\downarrow}=m_{BaSO_4}=233.0,05=11,65\left(g\right)\)

Đặt V (l) là thể tích dung dịch bazo cần tìm.

Ta có: nH+=0,2.0,1+0,1.0,1=0,03 (mol)nH+=0,2.0,1+0,1.0,1=0,03 (mol)

Mà Vaxit=100 ml⇒nH+=0,01 molVaxit=100 ml⇒nH+=0,01 mol

+) nOH−=V.0,02.2=0,04V molnOH−=V.0,02.2=0,04V mol

 (vì có 2 nhóm $O{}^{}$ trong dd $Ba(OH{}^{}$).

Ta có phương trình ion thu gọn sau:

${}^{}+O{}^{}\to {}^{}O$

Theo PTHH: nOH−=nH+⇒0,04V=0,01

100 ml dd A có \(n_{H^+}=\dfrac{0,1.0,2+0,1.0,1}{3}=0,01\left(mol\right)\).

Để trung hòa thì cần \(n_{OH^-}=n_{H^+}=0,01\left(mol\right)\).

\(\Leftrightarrow0,02.2.V_{Ba\left(OH\right)_2}=0,01\)

\(\Leftrightarrow V_{Ba\left(OH\right)_2}=0,25\)

1 Để trung hòa \(\Leftrightarrow\)n_OH^-=n_H⁺

    n_H⁺=V_A*\(C_M\)=0,1*(0,1+0,1)=0,02mol

   \(\Rightarrow\)n_\(OH^-\)=0,02 mol

   V=\(\dfrac{2n_{OH^-}}{C_M}\)(do n_\(OH^-\)=2n_{\(Ba\left(OH\right)_2\)})

     =\(\dfrac{2\cdot0,02}{0,02}\)=2l

Bài 6:

\(2M+3Cl_2\rightarrow2MCl_3\\ m_{Cl_2}=m_{MCl_3}-m_M=32,5-11,2=21,3\left(g\right)\\ n_{Cl_2}=\dfrac{21,3}{71}=0,3\left(mol\right)\\ n_M=\dfrac{2}{3}.n_{Cl_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ M_M=\dfrac{m_M}{n_M}=\dfrac{11,2}{0,2}=56\left(\dfrac{g}{mol}\right)\\ \Rightarrow M\left(III\right):Sắt\left(Fe=56\right)\)

Bài 5:

\(KL:X\left(II\right)\\ X+2HCl\rightarrow XCl_2+H_2\\ n_{HCl}=0,4.0,3=0,12\left(mol\right)\\ n_X=n_{XCl_2}=n_{H_2}=\dfrac{0,12}{2}=0,06\left(mol\right)\\ a,M_X=\dfrac{3,36}{0,06}=56\left(\dfrac{g}{mol}\right)\Rightarrow X\left(II\right):Sắt\left(Fe=56\right)\\ b,V_{ddFeCl_2}=V_{ddHCl}=0,4\left(l\right)\\ C_{MddFeCl_2}=\dfrac{0,06}{0,4}=0,15\left(M\right)\)

Bài 4:

\(KL:A\left(III\right)\\ 2A+3H_2SO_4\rightarrow A_2SO_4+3H_2\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\\ n_A=\dfrac{2}{3}.0,15=0,1\left(mol\right)\\ a,M_A=\dfrac{2,7}{0,1}=27\left(\dfrac{g}{mol}\right)\\ \Rightarrow A\left(III\right):Nhôm\left(Al=27\right)\\ b,m_{ddH_2SO_4}=\dfrac{0,15.98.100}{9,8}=150\left(g\right)\)

\(a.n_{NaOH}=1.0,15=0,15\left(mol\right)\\ n_{KOH}=0,5.0,1=0,05\left(mol\right)\\ \left[Na^+\right]=\left[NaOH\right]=\dfrac{0,15}{0,15+0,1}=0,6\left(M\right)\\ \left[K^+\right]=\left[KOH\right]=\dfrac{0,05}{0,1+0,15}=0,2\left(M\right)\\ \left[OH^-\right]=0,2+0,6=0,8\left(M\right)\\ b.2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4}=\dfrac{1}{2}.\left(n_{KOH}+n_{NaOH}\right)=\dfrac{0,15+0,05}{2}=0,1\left(mol\right)\\ a=C_{MddH_2SO_4}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

\(a)n_{Ba\left(OH\right)_2}=0,05\cdot0,2\cdot2=0,02mol\\ pH=1\Rightarrow\left[OH^-\right]=0,1M\Rightarrow n_{HCl}=0,1\cdot0,3=0,03mol\\ n_{Ba\left(OH\right)_2}+n_{HCl}=0,02+0,03=0,05mol\\ \Rightarrow C_M=\dfrac{0,05}{0,5}=0,1M\Rightarrow pH=1\)

\(n_{Mg}=\dfrac{24}{24}=1\left(mol\right)\)

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

\(1.........2............................1\)

\(V_{H_2}=1\cdot22.4=22.4\left(l\right)\)

\(m_{dd_{HCl}}=\dfrac{2\cdot36.5}{7.3\%}=1000\left(g\right)\)

\(\left[Na^+\right]=\dfrac{2.0,015}{2.0,015}=1M\)

\(\left[OH^-\right]=\dfrac{2.0,015}{2.0,015}=1M\)

\(\left[H^+\right]=\dfrac{2.1,5.0,015}{2.0,015}=1,5M\)

\(\left[SO_4^{2-}\right]=\dfrac{1,5.0,015}{2.0,015}=0,75M\)

a ) Gọi x là số mol NaOH , y là số mol KOH
Ta có : 15,2 g hỗn hợp NaOH va KOH
=> Ta có phương trình : 40x + 56y = 15,2 . . . . . ( 1 )

Ta có phương trình phản ứng :

NaOH + HCl ———→ NaCl + H2O
. x ———————–→ x mol

KOH + HCl ———→ KCl + H2O
. y ———————→ y mol

Thu được 20,75 g các muối Clorua
=> Ta có phương trình : 58,5x + 74.5y = 20,75 . . . . .( 2 )

Từ ( 1 ) và ( 2 ) ta có hệ 2 phương trình : Giải hệ ta có : x = 0,1 mol , y = 0,2 mol

x = 0,1 mol => n NaOH = 0,1 mol => m NaOH = 4 gam
y = 0,2 mol => n KOH = 0,1 mol => m KOH = 11,2 gam

% khối lượng mỗi chất :

. . . . . . . . . m NaOH tan x 100 . . . . 4 x 100
% NaOH = ————————— = ————— = 26,32 %
. . . . . . . . . . . m hỗn hợp . . . . . . . . .15,2

% KOH = 100 % – % NaOH = 100 % – 26,32 % = 73,68 %

Σ n HCl đã phản ứng = 0,1 + 0,2 = 0,3 mol => m HCl tan = 10,95 gam

. . . . . . . . . . . . .m HCl tan x 100 . . . .10,95 x 100 . . . .10,95
=> C% HCl = ————————— = ——————– = ———– = 5,475 % ≈ 5,5 %
. . . . . . . . . . . . . . . .m dd HCl . . . . . . . . 200 . . . . . . . .2

Vậy % NaOH = 26,32 % ; % KOH = 73,68 % ; % HCl phản ứng = 5,5%

hihi tự sướng