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PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
Ta có: \(n_{CuCl_2}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}\\n_{NaOH}=0,4\left(mol\right)=n_{NaCl}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\m_{CuO}=0,2\cdot80=16\left(g\right)\\m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\\C\%_{NaOH}=\dfrac{0,4\cdot40}{200}\cdot100\%=8\%\end{matrix}\right.\)
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=0,3.56326.100%=5,32%0,3.56326.100%=5,32%
nKCl=2nZnCl2=0,3(mol)KCl=2nZnCl2=0,3(mol)
C% KCl=0,3.74,5316.100%=7,07%
\(n_{NaOH}=\dfrac{200.10\%}{40}=0,5\left(mol\right)\)
\(PTHH:CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)
bđ: 0,3 0,5
pứ: 0,25 0,5 0,5 0,25
[ ]: 0,05 0 0,5 0,25
\(PTHH:Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
(mol) 0,25 0,25
\(a.m_C=80.0,25=20\left(g\right)\)
\(b.m_{NaCl}=58,5.0,5=29,25\left(g\right)\\ m_{Cu\left(OH\right)_2}=0,25.98=24,5\left(g\right)\\ m_{CuCl_2\left(du\right)}=135.0,05=6,75\left(g\right)\)
\(c.m_{ddspu}=100+200-24,5=275,5\left(g\right)\\ C\%_{ddCuCl_2\left(du\right)}=\dfrac{135.0,05}{275,5}.100=2,45\left(\%\right)\\ C\%_{ddNaCl}=\dfrac{0,5.58,5}{275,25}.100=10,62\left(\%\right)\)
nCuCl2 = \(\dfrac{270.15\%}{100\%.135}\)= 0,3(mol)
CuCl2 + 2KOH ➝ Cu(OH)2↓ + 2KCl
0,3 ➝ 0,6 ➝ 0,3 ➝ 0,6 (mol)
a, mCu(OH)2 = 0,3.98= 29,4(g)
b, m dd KOH = \(\dfrac{0,6.56.100\%}{20\%}\)= 168(g)
c, mKCl = 0,6.74,5 = 44,7(g)(*)
Áp dụng định luật bảo toàn khối lượng:
=> mdd KCl= mCuCl2 + m dd KOH - mCu(OH)2
⇔ mdd 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.CuCl2 + 2NaOH -> Cu(OH)2 + 2NaCl
0.15 0.3 0.15 0.3
Cu(OH)2 -> CuO + H2O
0.15 0.15
nNaOH = 0.3 mol
\(CM_{CuCl2}=\dfrac{0.15}{2}=0.075M\)
b.Vdd sau phản ứng = 0.2 + 0.15 = 0.35l
\(CM_{NaCl}=\dfrac{0.3}{0.35}=0.86M\)
c.mCuO = \(0.15\times80=12g\)
\(a,PTHH:CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\\ ...0,2......0,4.......0,2........0,4\left(mol\right)\\ b,n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\\ m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\ c,m_{KOH}=0,4\cdot56=22,4\left(g\right)\\ m_{dd_{KOH}}=\dfrac{22,4\cdot100\%}{20\%}=112\left(g\right)\\ m_{dd_{KCl}}=m_{CuCl_2}+m_{dd_{KOH}}-m_{Cu\left(OH\right)_2}=27+112-19,6=119,4\left(g\right)\)
\(d,C\%_{dd_{KCl}}=\dfrac{74,5\cdot0,4}{119,4}\cdot100\%\approx24,96\%\)