a, \(CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\)

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

b, \(n_{CuCl_2}=\dfrac{20,25}{135}=0,15\left(mol\right)\)

Theo PT: \(n_{KOH}=2n_{CuCl_2}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,3}{0,3}=1\left(M\right)\)

c, \(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,15\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,15.80=12\left(g\right)\)

nZnCl2 =40,8/136=0,3mol

nNaOH= 0,1.0,5=0,05mol

a)

pt : ZnCl2 + 2NaOH ------> Zn(OH)2\(\downarrow\) + 2NaCl

ncó: 0,3 0,05

n pứ: 0,025<------0,05-------->0,025-------->0,05

n dư: 0,275 0

b)

mZnCl2 dư = 0,275.136=37,4g

mNaCl=0,05.58,5=2,925g

c)

pt : Zn(OH)2 ---to--> ZnO + H2O

n pứ : 0,025------------>0,025

mZnO=0,025.81=2,025g

d)

vdd sau pứ =Vdd NaOH =0,1l

CM(ZnCl2 dư )=0,025/0,1=0,25M

CM(NaOH)=0,05/0,1= 0,5M

a) PTHH: \(3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\)

b) Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,5\cdot0,1=0,05\left(mol\right)\\n_{FeCl_3}=0,2\cdot0,2=0,04\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,05}{3}< \dfrac{0,04}{1}\) \(\Rightarrow\) NaOH p/ứ hết, FeCl₃ còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,05\left(mol\right)\\n_{FeCl_3\left(dư\right)}=\dfrac{7}{300}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,05\cdot58,5=2,925\left(g\right)\\m_{FeCl_3\left(dư\right)}=\dfrac{7}{300}\cdot162,5\approx3,8\left(g\right)\end{matrix}\right.\)

c) PTHH: \(2Fe\left(OH\right)_3\xrightarrow[]{t^o}Fe_2O_3+3H_2O\)

Ta có: \(n_{Fe\left(OH\right)_3}=\dfrac{1}{60}\left(mol\right)\) \(\Rightarrow n_{H_2O}=\dfrac{1}{40}\left(mol\right)\) \(\Rightarrow m_{H_2O}=\dfrac{1}{40}\cdot18=0,45\left(g\right)\)

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

a) \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\left(1\right)\)

     \(Cu\left(OH\right)_2\xrightarrow[t^o]{}CuO+H_2O\left(2\right)\)

b) \(Pt\left(1\right):n_{Cu\left(OH\right)2}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)

  \(Pt\left(2\right):n_{Cu\left(OH\right)2}=n_{CuO}=0,25\left(mol\right)\Rightarrow m_{Cu}=0,25.64=16\left(g\right)\)

c) Pt(1) : \(n_{NaOH}=n_{NaCl}=0,5\left(mol\right)\Rightarrow m_{NaCl}=0,5.58,5=29,25\left(g\right)\)

\(n_{CuCl2}=\frac{54}{135}=0.4mol\)

\(n_{NaOH}=\frac{40}{40}=1mol\)

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

\(0.4\) mol : 1 mol

Lập tỉ lệ: \(\frac{0.4}{1}< \frac{1}{2}\) ( NaOH dư, tính toán dựa vào CuCl\(_2\)

\(\rightarrow n_{CuCl2}=n_{Cu\left(OH\right)2}=0.4mol\)

\(n_{NaOH}=n_{NaCl}=2n_{CuCl2}=0.4\cdot2=0.8mol\)

\(\rightarrow n_{NaOHdư}=1-0.8=0.2mol\)

a, \(m_{Cu\left(OH\right)2}=0.4\cdot98=39.2\left(g\right)\)

\(m_{NaOHdư}=0.2\cdot40=8\left(g\right)\)

b, \(m_{NaCl}=0.4\cdot58.5=23.4\left(g\right)\)

\(n_{FeCl_3}=0.2\cdot0.4=0.08\left(mol\right)\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)

\(0.08...........0.24..............0.08\)

\(2Fe\left(OH\right)_3\underrightarrow{^{^{t^0}}}Fe_2O_3+3H_2O\)

\(0.08...........0.04\)

\(m_{Fe_2O_3}=0.04\cdot160=6.4\left(g\right)\)

\(V_{dd_{NaOH}}=\dfrac{0.24}{0.5}=0.48\left(l\right)\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\) (1)

\(2Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\) (2)

\(n_{FeCl_3}=0,2.0,4=0,08\left(mol\right)\)

Bảo toàn nguyên tố Fe : \(n_{FeCl_3}=2n_{Fe_2O_3}=0,08\left(mol\right)\)

=> \(n_{Fe_2O_3}=0,04\left(mol\right)\)

=> \(m_{Fe_2O_3}=0,04.160=6,4\left(g\right)\)

Theo PT (1) : \(n_{NaOH}=3n_{FeCl_3}=0,08.3=0,24\left(mol\right)\)

=> \(V_{NaOH}=\dfrac{0,24}{0,5}=0,48\left(l\right)\)

\(n_{FeCl_3}=C_M.V=0,4.0,2=0,08\left(mol\right)\)

PT: FeCl₃ + 3NaOH \(\rightarrow\) 3NaCl + Fe(OH)₃ \(\downarrow\)

       0,08 -> 0,24                     ->  0,08          (mol)

      2Fe(OH)₃ \(\underrightarrow{t^o}\) Fe₂O₃ + 3H₂O

        0,08       ->    0,04                                  (mol)

m_cr=m_Fe2O3=n.M=0,04.160=6,4(g)

\(V_{NaOH}=\dfrac{n}{C_M}=\dfrac{0,24}{0,5}=0,48\left(l\right)\)

a)

\(FeSO_4 + 2NaOH \to Fe(OH)_2 + Na_2SO_4\)

b)

\(n_{FeSO_4} = 0,4.0,5 = 0,2(mol) ; n_{NaOH} = 0,5.0,5 = 0,25(mol)\)

Ta thấy : \(2n_{FeSO_4} = 0,4 > n_{NaOH} = 0,25\) nên FeSO₄ dư.

Theo PTHH : 

\(n_{Fe(OH)_2} = 0,5n_{NaOH} = 0,125(mol)\\ \Rightarrow m_{Fe(OH)_2} = 0,125.90 = 11,25(gam)\)

c)

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

Theo PTHH : 

\(n_{Fe_2O_3} = 0,5n_{Fe(OH)_2} = 0,0625(mol)\\ \Rightarrow m_{Fe_2O_3} = 0,0625.160 = 10(gam)\)