nCuSO4=0,5.0,4=0,2 mol
CuSO4 +2NaOH=> Cu(OH)2+Na2SO4
0,2 mol                =>0,2 mol
Cu(OH)2=> CuO+H2O
0,2 mol =>0,2 mol
kết tủa A là Cu(OH)2 m=98.0,2=19,6g
cr B là CuO m=0,2.80=16g

a) \(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

PTHH: CuCl₂ + 2NaOH --> Cu(OH)₂ + 2NaCl

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\) => CuCl₂ hết, NaOH dư

PTHH: CuCl₂ + 2NaOH --> Cu(OH)₂ + 2NaCl

              0,2------>0,4-------->0,2------->0,4

            Cu(OH)₂ --t^o--> CuO + H₂O

               0,2-------------->0,2

=> m_CuO = 0,2.80 = 16(g)

b)

\(\left\{{}\begin{matrix}m_{NaOH\left(dư\right)}=20-0,4.40=4\left(g\right)\\m_{NaCl}=0,4.58,5=23,4\left(g\right)\end{matrix}\right.\)

\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)

PT: CuSO₄ + 2NaOH → Cu(OH)₂ + Na₂SO₄

mol 0,02 0,04 ← 0,02 → 0,02

2Cu(OH)₂ + O₂ \(\underrightarrow{t^o}\) 2CuO + 2H₂O

mol 0,02 0,01 ← 0,02 → 0,02

a) \(m_{Cu\left(OH\right)_2}=0,02.98=1,96\left(g\right)\)

b) V_{dung dịch sau phản ứng} = 200 + 300 = 500 (ml) = 0,5 (l)

\(C_{MNa_2SO_4}=\dfrac{0,02}{0,5}=0,04M\)

\(n_{CuSO_4}=0.2\cdot0.5=0.1\left(mol\right)\)

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

\(0.1.............0.2.................0.1..........0.1\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.1}{0.3+0.2}=0.2\left(M\right)\)

\(Cu\left(OH\right)_2\underrightarrow{^{^{t^0}}}CuO+H_2O\)

\(0.1.............0.1\)

\(m_{CuO}=0.1\cdot80=8\left(g\right)\)

Nồng độ của Na₂SO₄ mà em !

a)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)

\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow3Na_2SO_4+2Al\left(OH\right)_3\downarrow\)

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

\(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)

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

b) B gồm Fe(OH)₂, Cu(OH)₂

C gồm CuO, Fe₂O₃

a)\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)

\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow2Al\left(OH\right)_3+3Na_2SO_4\)

\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)

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

\(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)

b)B là các chất sau: \(Fe\left(OH\right)_2;Cu\left(OH\right)_2;Al\left(OH\right)_3\)

C là các chất sau: \(Fe_2O_3;CuO;Al_2O_3\) 

a) 3NaOH+FeCl3---->Fe(OH)3+3NaCl

b) n NaOH=0,15.2=0,3(mol)

Theo pthh

n FeCl3=1/3n NaOH=0,1(mol)

m dd FeCl3=\(\frac{162,5.100.0,1}{20}=81,25\left(g\right)\)

c) Theo pthh

n Fe(OH)3=1/3n NaOH=0,1(mol)

m Fe(OH)3=0,1.107=10,7(g)

d) 2Fe(OH)3--->Fe2O3+3H2O

Theo pthh

n Fe2O3=1/2n Fe(OH)3=0,05(mol)

m Fe2O3=0,05.160=8(g)

PTPU

Mg+ 2HCl\(\rightarrow\) MgCl₂+ H₂\(\uparrow\) (1)

MgCl₂+ 2NaOH\(\rightarrow\) Mg(OH)₂\(\downarrow\)+ 2NaCl (2)

Mg(OH)₂\(\xrightarrow[]{to}\) MgO+ H₂O (3)

có: n_Mg= \(\frac{4,8}{24}\)= 0,2( mol)

theo ptpư(1) có: n_HCl= 2n_Mg= 0,4( mol)

\(\Rightarrow\) m_{dd HCl}= \(\frac{0,4.36,5}{20\%}\)= 73( g)

có: n_MgCl2= n_H2= n_Mg= 0,2( mol)

\(\Rightarrow\) m_MgCl2= 0,2. 95= 19( g)

có: m_{dd sau pư}= m_Mg+ m_{dd HCl}- m_H2

= 4,8+ 73- 0,2. 2= 77,4( g)

\(\Rightarrow\) C%_MgCl2= \(\frac{19}{77,4}\). 100%= 24,55%

theo ptpư(2) có: n_Mg(OH)2= n_MgCl2= 0,2( mol)

\(\Rightarrow\) m_Mg(OH)2= 0,2. 58= 11,6( g)

theo ptpư(3) có: n_MgO= n_Mg(OH)2= 0,2( mol)

\(\Rightarrow\) m_MgO= 0,2. 40=8( g)