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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: \(n_{CuCl_2}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}\\n_{NaOH}=0,4\left(mol\right)=n_{NaCl}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\m_{CuO}=0,2\cdot80=16\left(g\right)\\m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\\C\%_{NaOH}=\dfrac{0,4\cdot40}{200}\cdot100\%=8\%\end{matrix}\right.\)
\(n_{NaOH}=\dfrac{200.10\%}{40}=0,5\left(mol\right)\)
\(PTHH:CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)
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
\(a.m_C=80.0,25=20\left(g\right)\)
\(b.m_{NaCl}=58,5.0,5=29,25\left(g\right)\\ m_{Cu\left(OH\right)_2}=0,25.98=24,5\left(g\right)\\ m_{CuCl_2\left(du\right)}=135.0,05=6,75\left(g\right)\)
\(c.m_{ddspu}=100+200-24,5=275,5\left(g\right)\\ C\%_{ddCuCl_2\left(du\right)}=\dfrac{135.0,05}{275,5}.100=2,45\left(\%\right)\\ C\%_{ddNaCl}=\dfrac{0,5.58,5}{275,25}.100=10,62\left(\%\right)\)
a) \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
b) \(n_{CuSO_4}=0,4.1=0,4\left(mol\right)\)
\(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,4}{1}< \dfrac{1}{2}\) => CuSO4 hết, NaOH dư
PTHH: CuSO4 + 2NaOH --> Cu(OH)2 + Na2SO4
_______0,4----->0,8---------->0,4-------->0,4
Cu(OH)2 --to--> CuO + H2O
0,4------------->0,4
=> mCuO = 0,4.80 = 32 (g)
c) \(\left\{{}\begin{matrix}m_{NaOH\left(dư\right)}=\left(1-0,8\right).40=8\left(g\right)\\m_{Na_2SO_4}=0,4.142=56,8\left(g\right)\end{matrix}\right.\)
Ta có \(n_{CuCl_2}=\frac{1,35}{135}=0,01\left(mol\right);n_{KOH}=\frac{2,8}{56}=0,05\left(mol\right)\)
PTHH \(CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2\downarrow+2KCl\) (1)
Thấy \(n_{CuCl_2pu}:n_{KOHpu}=\frac{0,01}{1}< \frac{0,05}{2}\Rightarrow CuCl_2hết\)tính theo \(CuCl_2\)
a) \(Cu\left(OH\right)_2-t^o\rightarrow CuO+H_2O\) (2)
\(\Rightarrow\) Chất rắn sau khi nung là CuO
Theo (1) : \(n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,01\left(mol\right)\)
Theo (2) : \(n_{CuO}=n_{Cu\left(OH\right)_2}=0,01\left(mol\right)\)
\(\Rightarrow m_{CuO}=80\times0,01=0,8\left(g\right)\)
b) Chất có trong dd sau phản ứng là KOH
Theo (1) \(n_{KOHpu}=2n_{CuCl_2}=0,02\left(mol\right)\)
\(\Rightarrow n_{KOHdu}=0,05-0,02=0,03\left(mol\right)\)
\(\Rightarrow m_{KOHdu}=0,03\times56=1,68\left(g\right)\)
\(n_{CuSO_4}=\dfrac{200.16}{160.100}=0,2mol\)
\(n_{NaOH}=\dfrac{200.10}{40.100}=0,5mol\)
CuSO4+2NaOH\(\rightarrow\)Cu(OH)2\(\downarrow\)+Na2SO4
-Ta có tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\rightarrow\)CuSO4 hết, NaOH dư.
Cu(OH)2\(\overset{t^0}{\rightarrow}\)CuO+H2O
\(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,2mol\)
a=\(m_{CuO}=0,2.80=16gam\)
\(m_{Cu\left(OH\right)_2}=0,2.98=19,6gam\)
\(n_{NaOH\left(pu\right)}=2n_{CuSO_4}=0,4mol\rightarrow n_{NaOH\left(dư\right)}=0,5-0,4=0,1mol\)
\(m_{NaOH\left(dư\right)}=0,1.40=4gam\)
\(n_{Na_2SO_4}=n_{CuSO_4}=0,2mol\rightarrow m_{Na_2SO_4}=0,2.136=27,2gam\)
\(m_{dd}=200+200-19,6=380,4gam\)
C%NaOH=\(\dfrac{4.100}{380,4}\approx1,05\%\)
C%Na2SO4=\(\dfrac{27,2.100}{380,4}\approx7,15\%\)
PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{CuCl_2}=0,2\cdot2=0,4\left(mol\right)\\n_{NaOH}=0,2\cdot2=0,4\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,4}{2}\) \(\Rightarrow\) CuCl2 còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,4\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}=n_{CuCl_2\left(dư\right)}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2\cdot80=16\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,4}{0,2+0,2}=1\left(M\right)\\C_{M_{CuCl_2}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\\\end{matrix}\right.\)
CuCl2 + 2NaOH -> Cu(OH)2 + 2NaCl (1)
Cu(OH)2 -> CuO + H2O (2)
Theo PTHH 1 và 2 ta có:
nCuCl2=nCu(OH)2=nCuO=2(mol)
nNaCl=nNaOH=2nCuCl2=4(mol)
mNaOH còn lại=200-4.40=40(g)
mCuO=80.2=160(g)
nNaCl=58,5.4=234(g)