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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)
$NaOH + SO_2 \to NaHSO_3$
$n_{NaOH} = n_{SO_2} = 0,1(mol)$
$\Rightarrow V_{dd\ NaOH} = \dfrac{0,1}{1} = 0,1(lít) = 100(ml)$
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: \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(l\right)=100\left(ml\right)\)
\(a.n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\\ Na_2O+H_2O\rightarrow NaOH\\ m_{ddNaOH}=193,8+6,2=200\left(g\right)\\C\%_{ddX}=C\%_{ddNaOH}=\dfrac{0,1.2.40}{200}.100=4\%\\ b.2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ a=m_{Cu\left(OH\right)_2}=\dfrac{0,2}{2}.98=9,8\left(g\right)\\ c.Cu\left(OH\right)_2\underrightarrow{to}CuO+H_2O\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{HCl}=2.n_{CuO}=2.n_{Cu\left(OH\right)_2}=2.0,1=0,2\left(mol\right)\\ V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(lít\right)=100\left(ml\right)\)
Pư: Mg+2HCl -> MgCl2 +H2
MgO+2HCl-> MgCl2 +H2O
MgCl2+2NaOH-> Mg(OH)2↓+2NaCl
Mg(OH)2-> MgO+H2O
nMgO=14/40=0.35 mol
-> nMg(OH)2=nMgCl2=nMgO=0.35 mol
-> nMgO=nMg=nMgCl2=0.35 mol
gọi x,y lần lượt là số mol của Mg,MgO
->mhh=24x+40y=2 (1)
nMgCl2=x+y=0.35 (2)
giải hệ (1)(2) có x=0.103125
y=0.246875
- khối lượng của bạn nhập sai vì klg hỗn hợp quá nhỏ
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)
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: \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(l\right)=100\left(ml\right)\)
nếu a cho là 2(g) thì còn đúng ko bạn