c) Cu(OH)2 ---to-> CuO + H2O;

ta có: nCu(OH)2=0,0775(mol)=> nCuO=0,0775(mol)

CuO + 2 HCl --> CuCl2 + H2O;

0,0775---0,155 (mol)

nHCl=0,155(mol)=>mHCl=0,155*36,5=5,6575(g)

a. mdd X= 6,2+193,8= 200(g)=> C%X=6,2⋅100200=3,1%6,2⋅100200=3,1%

b. 2NaOH + CuSO4 ---> Na2SO4 + Cu(OH)2;

0,155--------0,0775------------0,0775----------0,0775 (mol)

Ta có: nCuSO4=200⋅16100⋅160=0,2(mol)200⋅16100⋅160=0,2(mol)

nNaOH=6,2400,155(mol)6,2400,155(mol)

Xét tỉ lệ:nNaOHnNaOHpt=0,1552<nCuSO4nCuSO4pt=0,21nNaOHnNaOHpt=0,1552<nCuSO4nCuSO4pt=0,21

=> CuSO4 dư. Sản phẩm tính theo NaOH.

=> nNa2SO4=0,155/2= 0,0775(mol)=> mNa2SO4=0,0775*142=11,005(g).

nCu(OH)2=0,155/2=0,0775(mol)=> mCu(OH)2=0,0775*98=7,595(g).

=> mdd sau pư= 200+200-7,595=392,405(g)

=> C%ddA=11,005⋅100392,405=2,8%

a. m_dd X= 6,2+193,8= 200(g) ⇒C%X=\(\dfrac{6,2}{200}.100=31\%\)

b.\(n_{NaOH}=\dfrac{6,2}{40}=0,155\left(mol\right)\)

n\(_{CuSO_4}=\dfrac{200.16}{100.160}=0,2\left(mol\right)\)

PTHH: 2NaOH + CuSO₄ → Na₂SO₄ + Cu(OH)₂

_____0,155______0,0775___0,0775_____0,0775 (mol)

Ta có tỉ lệ:\(\dfrac{n_{NaOH}}{2}:\dfrac{n_{CuSO_4}}{1}=\dfrac{0,155}{2}< \dfrac{0,2}{1}\)

⇒NaOH pư hết, CuSO₄ pư dư.

\(m_{Na_2SO_4}=0,0775.142=11,005\left(g\right)\)

\(m_{Cu\left(OH\right)_2}=0,775.98=7,595\left(g\right)\)

\(\Rightarrow m_{dd}saupư=200+200-7,595=392,405\left(g\right)\)

\(\Rightarrow C\%_A=\dfrac{11,005}{392,405}.100=2,8\%\)

c. PTHH: Cu(OH)₂ \(\underrightarrow{t^o}\) CuO + H₂O

n\(_{Cu\left(OH\right)_2}\)=0,0775(mol)⇒ n_CuO=0,0775(mol)

PTHH: CuO + 2HCl→CuCl₂ + H₂O

0,0775_0,155 (mol)

n_HCl=0,155(mol)⇒m_HCl=0,155.36,5=5,6575(g)

\(a.n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\\ Na_2O+H_2O\rightarrow NaOH\\ m_{ddNaOH}=193,8+6,2=200\left(g\right)\\C\%_{ddX}=C\%_{ddNaOH}=\dfrac{0,1.2.40}{200}.100=4\%\\ b.2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ a=m_{Cu\left(OH\right)_2}=\dfrac{0,2}{2}.98=9,8\left(g\right)\\ c.Cu\left(OH\right)_2\underrightarrow{to}CuO+H_2O\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{HCl}=2.n_{CuO}=2.n_{Cu\left(OH\right)_2}=2.0,1=0,2\left(mol\right)\\ V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(lít\right)=100\left(ml\right)\)

PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

            \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

            \(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)

a) Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\) \(\Rightarrow C\%_{NaOH}=\dfrac{0,2\cdot40}{6,2+193,8}\cdot100\%=4\%\)

b) Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{CuSO_4}=\dfrac{200\cdot16\%}{160}=0,2\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) \(\Rightarrow\) CuSO₄ còn dư, tính theo NaOH

\(\Rightarrow n_{Cu\left(OH\right)_2}=0,1\left(mol\right)=n_{CuO}\) \(\Rightarrow m_{CuO}=0,1\cdot80=8\left(g\right)\)

c) PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

Theo PTHH: \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(l\right)=100\left(ml\right)\)

gọi số mol của NaOH = x , Ba(OH)2 = y . số mol OH- = x + 2y

pt : H+ + OH- → H2O 

nH+ =2. 0,035. 2=0,14 mol = x+ 2y 

khối lượng kết tủa = m_SO42-  + m_Ba2+ = 96.0.07 + 137.y= 9,32 =>y=0,02 => x= 0,1

nồng độ tự tính.