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

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

b, \(n_{MgCl_2}=0,2.0,25=0,05\left(mol\right)\)

 Theo PT: \(n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0,05\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,05.40=2\left(g\right)\)

c, \(n_{NaOH}=2n_{MgCl_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{ddNaOH}=\dfrac{0,1.40}{15\%}=\dfrac{80}{3}\left(g\right)\)

a, \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

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

b, \(n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\)

Theo PT: \(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

c, \(n_{NaOH}=2n_{CuCl_2}=0,4\left(mol\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,4.40}{200}.100\%=8\%\)

PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)

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

Ta có: \(n_{CuCl_2}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}\\n_{NaOH}=0,4\left(mol\right)=n_{NaCl}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\m_{CuO}=0,2\cdot80=16\left(g\right)\\m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\\C\%_{NaOH}=\dfrac{0,4\cdot40}{200}\cdot100\%=8\%\end{matrix}\right.\)

gọi a, b lần lượt là số mol MgCl₂ và CuCl₂ trong 25g hỗn hợp

ta có: n_AgCl= \(\dfrac{17,22}{143,5}\)= 0,12( mol)

PTPU

MgCl₂+ 2AgNO₃\(\rightarrow\) Mg(NO₃)₂+ 2AgCl\(\downarrow\)

..a......................................................2a...... mol

CuCl₂+ 2AgNO₃\(\rightarrow\) Cu(NO₃)₂+ 2AgCl\(\downarrow\)

..b...................................................2b...... mol

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

..a.............................a.............................. mol

CuCl₂+ 2NaOH\(\rightarrow\) Cu(OH)₂+ 2NaCl

..b.............................b......................... mol

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

..a....................a................. mol

Cu(OH)₂\(\xrightarrow[]{to}\) CuO+ H₂O

..b...................b............... mol

ta có hệ pt

\(\left\{{}\begin{matrix}2a+2b=0,12\\40a+80b=3,8\end{matrix}\right.\)

\(\Rightarrow\) \(\left\{{}\begin{matrix}a=0,025\\b=0,035\end{matrix}\right.\)

\(\Rightarrow\) C%_MgCl2= \(\dfrac{0,025.95}{25}\).100%= 9,5%

C%_CuCl2= \(\dfrac{0,035.135}{25}\). 100%= 18,9%

C → + O 2 A C O C O 2 → + F e O , t 0 B : C O 2 → + C a ( O H ) 2 K : C a C O 3 D :   C a H C O 3 2 C F e F e O → + H C l H 2 E : F e C l 2 → + N a O H F : : F e O H 2 → t 0 , k k G : F e 2 O 3

⇒ Chọn A.

Mg + HCl - MgCl₂ + H₂

Al + HCl - AlCl3 + H₂

còn Cu đứng sau H nên không phản ứng với HCl.

n_CuO = \(\dfrac{16}{80}=0,2\) mol

Pt: 2Cu + O₂ --to--> 2CuO

0,2 mol<--------------0,2 mol

.....Mg + 2HCl --> MgCl₂ + H₂

0,2 mol<-----------0,2 mol

.....2Al + 6HCl --> 2AlCl₃ + 3H₂

....MgCl₂ + 2NaOH --> Mg(OH)₂ + 2NaCl

...0,2 mol<---------------0,2 mol

....AlCl₃ + 3NaOH --> Al(OH)₃ + 3NaCl

....Al(OH)₃ + NaOH --> NaAlO₂ + 2H₂O

..........................................(tan)

...Mg(OH)2 --to--> MgO + H2O

0,2 mol<------------ 0,2 mol

n_MgO = \(\dfrac{8}{40}=0,2\) mol

m_Cu = 0,2 . 64 = 12,8 (g)

m_Mg = 0,2 . 24 = 4,8 (g)

m_Al = m_hh - m_Cu - m_Mg = 20 - 12,8 - 4,8 = 2,4 (g)

% m_Cu = \(\dfrac{12,8}{20}.100\%=64\%\)

% m_Mg = \(\dfrac{4,8}{20}.100\%=24\%\)

% m_Al = \(\dfrac{2,4}{20}.100\%=12\%\)

\(n_{NaOH}=\dfrac{200.15\%}{40}=0,75\left(mol\right)\)

\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,0001V\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=0,00005V\left(mol\right)\end{matrix}\right.\)

PTHH: 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

          0,0002V<-0,0001V

           6NaOH + Fe₂(SO₄)₃ --> 3Na₂SO₄ + 2Fe(OH)₃

         0,0003V<-0,00005V---------------->0,0001V

=> 0,0002V + 0,0003V = 0,75

=> V = 1500 (ml)

n_Fe(OH)3 = 0,15 (mol)

=> m₁ = 0,15.107 = 16,05 (g)

PTHH: 2Fe(OH)₃ --t^o--> Fe₂O₃ + 3H₂O

              0,15--------->0,075

=> m_Fe2O3 = 0,075.160 = 12 (g)