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PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
a) Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,2\cdot40}{193,8+6,2}\cdot100\%=4\%\)
b) Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{CuSO_4}=\dfrac{200\cdot16\%}{160}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) \(\Rightarrow\) CuSO4 còn dư, NaOH p/ứ hết
\(\Rightarrow n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=n_{CuO}=0,1\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,1\cdot80=8\left(g\right)\)
c) PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Theo PTHH: \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(l\right)=100\left(ml\right)\)
a) Na2O +H2O-->2NaOH (1)
2NaOH +CuSO4 -->Na2SO4+ Cu(OH)2 (2)
Cu(OH)2 -to-> CuO +H2O (3)
b) mNa2O=8.100/100=8(g)
=>nNa2O=8/62=0,13(mol)
theo(2) :nCu(OH)2=1/2nNaOH=0,065(mol)
theo(3):nCuO=nCu(OH)2=0,065(mol)
=>mCuO=0,065.80=5,2(g)
c) CuO +2HCl-->CuCl2+H2O (4)
theo (4) : nHCl=2nCuO=0,13(mol)
mddHCl 25%=0,13.36,5.100250,13.36,5.10025=18,98(g)
Câu 1:
PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
a) Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{12,4}{62}=0,4\left(mol\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,4\cdot40}{12,4+193,8}\cdot100\%\approx7,76\%\)
b) Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,4\left(mol\right)\\n_{CuSO_4}=\dfrac{100\cdot16\%}{160}=0,1\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,4}{2}>\dfrac{0,1}{1}\) \(\Rightarrow\) NaOH còn dư, CuSO4 p/ứ hết
\(\Rightarrow n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,1\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,1\cdot80=8\left(g\right)\)
Bài 2 :
a)
$Cu + 2H_2SO_{4_{đặc}} \to CuSO_4 + SO_2 + 2H_2O$
$n_{Cu} = n_{SO_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$\%m_{Cu} = \dfrac{0,1.64}{15}.100\% = 42,67\%$
$\%m_{CuO} = 100\% -42,67\% = 57,33\%$
b)
$NaOH + SO_2 \to NaHSO_3$
$n_{NaOH} = n_{SO_2} = 0,1(mol)$
$\Rightarrow V_{dd\ NaOH} = \dfrac{0,1}{1} = 0,1(lít) = 100(ml)$
a) mNaOH= 4%. 100= 4(g)
mCuSO4= 16%. 200= 32(g)
=> nNaOH= 4/40= 0,1(mol)
nCuSO4= 32/160= 0,2(mol)
PTHH: 2 NaOH + CuSO4 -> Cu(OH)2 + Na2SO4 (1)
Ta có: 0,1/2 < 0,2/1
=> NaOH hết, CuSO4 dư, tính theo nNaOH.
=> Kết tủa là Cu(OH)2
PTHH: (2) Cu(OH)2 -to-> CuO + H2O
n(rắn_đen)= nCuO = nCu(OH)2 (2)= nCu(OH)2 (1)= nNaOH/2= 0,1/2 0,05(mol)
=> a= m(rắn_đen)= mCuO= 0,05.80= 4(g)
b) dd A gồm dd CuSO4(dư) và dd Na2SO4
Ta có: nCuSO4(dư)= 0,2 - (0,1:2)= 0,15(mol)
=> mCuSO4(dư)= 0,15. 160= 24(g)
nNa2SO4= nNaOH/2= 0,1/2= 0,05(mol)
=> mNa2SO4= 142. 0,05= 7,1(g)
=> mddA= mddCuSO4 + mddNaOH - mCu(OH)2
<=> mddA= 200 + 100 - 0,05. 98= 295,1(g)
=> C%ddCuSO4(dư)= (24/295,1).100 \(\approx\) 8,133%
C%ddNa2SO4= (7,1/295,1).100 \(\approx\) 2,406%
c) PTHH: (3) CuO + 2 HCl -> CuCl2 + H2O
Ta có: nCuO(3)= nCuO(2)= 0,05(mol)
=> nHCl(3)= 2.0,05= 0,1(mol)
=> VddHCl= 0,1/ 2= 0,05(l)
Na2O+H2O->2NaOH
0,1 0,1 0,2
2NaOH+CuSO4->Na2SO4+Cu(OH)2
0,2 0,1 0,1 0,1
a.mNaOH=0,2.40=8(g)
mdd NaOH=6,2+193,8=200(g)
C%dd NaOH=8/200.100%=4%
b.mCu(OH)2=0,1.98=9,8(g)
c.Cu(OH)2->CuO+H2O
0,1 0,1 0,1
CuO+2HCl->CuCl2+H2O
0,1 0,2
VddHCl=0,2/2=0,1(l)
Câu c mình ko biết làm đúng hay ko
PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
a) Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\) \(\Rightarrow C\%_{NaOH}=\dfrac{0,2\cdot40}{6,2+193,8}\cdot100\%=4\%\)
b) Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{CuSO_4}=\dfrac{200\cdot16\%}{160}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) \(\Rightarrow\) CuSO4 còn dư, tính theo NaOH
\(\Rightarrow n_{Cu\left(OH\right)_2}=0,1\left(mol\right)=n_{CuO}\) \(\Rightarrow m_{CuO}=0,1\cdot80=8\left(g\right)\)
c) PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Theo PTHH: \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(l\right)=100\left(ml\right)\)