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

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

\(m_{ct_{CuCl_2}}=\frac{m_{dd}.C\%}{100\%}=\frac{270.15}{100}=40,5\left(gam\right)\)

\(n_{CuCl_2}=\frac{m}{M}=\frac{40,5}{135}=0.3\left(mol\right)\)

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

(mol)        1                     2                      1                        2

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

a) Khối lượng kết tủa thu được là:

\(m_{Cu\left(OH\right)_2}=n.M=0,3.98=29,4\left(gam\right)\)

b) Khối lượng KOH đã tham gia phản ứng là:

\(m_{KOH}=n.M=0,6.56=33,6\left(gam\right)\)

Khối lượng dung dịch KOh đac tham gia phản ứng là:

\(m_{dd_{KOH}}=\frac{m_{ct}}{C\%}.100\%=\frac{33,6}{20}.100=168\left(gam\right)\)

c) Khối lượng dung dịch sau phản ứng :

\(m_{dd}=(270+168)-29,4=408,6\left(gam\right)\)

\(m_{KCl}=n.M=0,6.74,5=44,7\left(gam\right)\)

Nồng đọ phần trăm của dung dịch sao phản ứng là:

\(C\%=\frac{m_{KCl}}{m_{dd}}.100\%=\frac{44,7}{408,6}.100\%\approx10,94\%\)

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

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

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

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

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

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