n_CuCl2 = \(\dfrac{270.15\%}{100\%.135}\)= 0,3(mol)

CuCl₂ + 2KOH ➝ Cu(OH)₂↓ + 2KCl

  0,3  ➝    0,6    ➝     0,3        ➝  0,6         (mol)

a, m_Cu(OH)2 = 0,3.98= 29,4(g)

b,  m _{dd KOH} = \(\dfrac{0,6.56.100\%}{20\%}\)= 168(g)

c, m_KCl = 0,6.74,5 = 44,7(g)(*)

   Áp dụng định luật bảo toàn khối lượng:

=> m_{dd KCl}= m_CuCl2 + m _{dd KOH} - m_Cu(OH)2

_⇔ m_{dd KCl} = 0,3.135+ 168 - 29,4 = 179,1(g)(**)

Từ (*) và (**) ⇒  C%_KCl = \(\dfrac{44,7}{179,1}\).100%\(\approx\) 24,96%

n_CuCl2 = \(\dfrac{270.15\%}{100\%.135}\)= 0,3(mol)

CuCl₂ + 2KOH ➝ Cu(OH)₂↓ + 2KCl

  0,3  ➝    0,6    ➝     0,3        ➝  0,6         (mol)

a, m_Cu(OH)2 = 0,3.98= 29,4(g)

b, m_KOH = 0,6.56 = 33,6(g)

c, m_KCl = 0,6.74,5 = 44,7(g)(*)

   m _{dd KOH} = \(\dfrac{33,6.100\%}{20\%}\)= 168(g)

Áp dụng định luật bảo toàn khối lượng:

=> m_{dd KCl}= m_CuCl2 + m _{dd KOH} - m_Cu(OH)2

_⇔ m_{dd KCl} = 0,3.135+ 168 - 29,4 = 179,1(g)(**)

Từ (*) và (**) ⇒  C%_KCl = \(\dfrac{44,7}{179,1}\).100%\(\approx\) 24,96%

PTHH: \(CuCl_2+2KOH\rightarrow2KCl+Cu\left(OH\right)_2\downarrow\)

a) Ta có: \(n_{CuCl_2}=\dfrac{270\cdot15\%}{135}=0,3\left(mol\right)=n_{Cu\left(OH\right)_2}\) \(\Rightarrow m_{Cu\left(OH\right)_2}=0,3\cdot98=29,4\left(g\right)\)

b) Theo PTHH: \(n_{KOH}=2n_{CuCl_2}=0,6mol\) \(\Rightarrow m_{ddKOH}=\dfrac{0,6\cdot57}{20\%}=171\left(g\right)\)

c) Theo PTHH: \(n_{KCl}=n_{KOH}=0,6mol\) \(\Rightarrow m_{KCl}=0,6\cdot74,5=44,7\left(g\right)\)

Mặt khác: \(m_{dd}=m_{ddCuCl_2}+m_{ddKOH}-m_{Cu\left(OH\right)_2}=411,6\left(g\right)\)

\(\Rightarrow C\%_{KCl}=\dfrac{44,7}{411,6}\cdot100\%\approx10,86\%\)

\(a,PTHH:CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\\ ...0,2......0,4.......0,2........0,4\left(mol\right)\\ b,n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\\ m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\ c,m_{KOH}=0,4\cdot56=22,4\left(g\right)\\ m_{dd_{KOH}}=\dfrac{22,4\cdot100\%}{20\%}=112\left(g\right)\\ m_{dd_{KCl}}=m_{CuCl_2}+m_{dd_{KOH}}-m_{Cu\left(OH\right)_2}=27+112-19,6=119,4\left(g\right)\)

\(d,C\%_{dd_{KCl}}=\dfrac{74,5\cdot0,4}{119,4}\cdot100\%\approx24,96\%\)

Đáp án A

a)PTHH: ZnCl2+2KOH---->Zn(OH)2+2KCl

b)

mZnCl2=204.10100=20,4(g)ZnCl2=204.10100=20,4(g)

nZnCl2=20,4136=0,15(mol)ZnCl2=20,4136=0,15(mol)

nKOH=112.20%56=0,4(mol)KOH=112.20%56=0,4(mol)

=> 0,15/1 <  0,4/1=> KOH dư

Theo pthh, ta có :

nCu(OH)2=nZnCl2=0,15(mol)Cu(OH)2=nZnCl2=0,15(mol)

mCu(OH)2=0,15.98=14,7(g)Cu(OH)2=0,15.98=14,7(g)

c) m dd sau pư=204+112=316(g)

Theo pthh

nKOH=2nZnCl2=0,3(mol)KOH=2nZnCl2=0,3(mol)

C% KOH=

nKCl=2nZnCl2=0,3(mol)KCl=2nZnCl2=0,3(mol)

C% KCl=

\(n_{KOH}=\dfrac{100.14}{100.56}=0,25(mol)\\ 2KOH+CuCl_2\to Cu(OH)_2\downarrow+2KCl\\ \Rightarrow n_{CuCl_2}=n_{Cu(OH)_2}=0,125(mol);n_{KCl}=0,25(mol)\\ a,m_{CuCl_2}=0,125.135=16,875(g)\\ b,m_{Cu(OH)_2}=0,125.98=12,25(g)\\ c,C\%_{KCl}=\dfrac{0,25.74,5}{100+16,875-12,25}.100\%=17,8\%\\ d,Cu(OH)_2\xrightarrow{t^o}CuO+H_2O\\ \Rightarrow n_{CuO}=0,125(mol)\\ \Rightarrow m_{CuO}=0,125.80=10(g)\)

Đáp án A

→ a =   0 , 0325

a, \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_{3\downarrow}\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

b, \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)

\(n_{KOH}=3n_{FeCl_3}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)