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a)
$Fe +H_2SO_4 \to FeSO_4 + H_2$
$FeSO_4 + 2KOH \to Fe(OH)_2 + K_2SO_4$
$4Fe(OH)_2 + O_2 \xrightarrow{t^o} 2Fe_2O_3 + 4H_2O$
$n_{Fe_2O_3} = \dfrac{20}{160} = 0,125(mol)$
Theo PTHH : $n_{Fe} = 2n_{Fe_2O_3} = 0,25(mol)$
$m_{Fe} = 0,25.56 = 14(gam)$
b)
$n_{H_2} = n_{Fe} = 0,25(mol)$
$V_{H_2} = 0,25.22,4 = 5,6(lít)$
c)
$n_{H_2SO_4} = n_{Fe} = 0,25(mol)$
$V_{dd\ H_2SO_4} = \dfrac{0,25}{1} = 0,25(lít) = 250(ml)$
\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ FeSO_4+2KOH\rightarrow Fe\left(OH\right)_2+K_2SO_4\\4 Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)
\(a.n_{Fe_2O_3}=\dfrac{20}{160}=0,125\left(mol\right)\\ n_{H_2}=n_{H_2SO_4}=n_{Fe}=n_{FeSO_4}=n_{Fe\left(OH\right)_2}=\dfrac{4}{2}.0,125=0,25\left(mol\right)\\ m_{Fe}=0,25.56=14\left(g\right)\\ b.V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\\ c.V_{ddH_2SO_4}=\dfrac{0,25}{1}=0,25\left(l\right)=250\left(ml\right)\)
a , \(nFe=\dfrac{11,2}{56}=0,2\left(mol\right)\)
, pthh:
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
1mol 2mol 1mol 1mol
0,2 0,4 0,2 0,2
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)
1mol 2mol 1mol 2mol
0,2 0,4 0,2 0,4
b, \(mFe\left(OH\right)_2=0,2.90=18\left(gam\right)\)
Oxit kim loại : R2On
\(\%R = \dfrac{2R}{2R + 16n}.100\% = 60\%\\ \Rightarrow R = 12n\)
Với n = 2 thì R = 24(Magie)
Vậy oxit là MgO
\(MgO+ H_2SO_4 \to MgSO_4 + H_2O\\ n_{MgSO_4} = n_{H_2SO_4} = n_{MgO} = \dfrac{20}{40} = 0,5(mol)\\ \Rightarrow m_{dd\ H_2SO_4} =\dfrac{0,5.98}{10\%} = 490(gam)\\ m_{dd\ sau\ pư} = 20 + 490 = 510(gam)\\ \Rightarrow C\%_{MgSO_4} = \dfrac{0,5.120}{510}.100\% = 11,76\%\)
a) \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
Theo PTHH: \(n_{Ca\left(OH\right)_2}=n_{CO_2}=0,1\left(mol\right)\)
\(V_{Ca\left(OH\right)_2}=200ml=0,2l\)
\(\Rightarrow C_{MCa\left(OH\right)_2}=\dfrac{n_{Ca\left(OH\right)_2}}{V_{Ca\left(OH\right)_2}}=\dfrac{0,1}{0,2}=0,5M\)
b) Theo PTHH có: \(n_{CaCO_3}=n_{CO_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=n_{CaCO_3}.M_{CaCO_3}=0,1.74=7,4\left(g\right)\)
c) Cu(OH)2 ---to-> CuO + H2O;
ta có: nCu(OH)2=0,0775(mol)=> nCuO=0,0775(mol)
CuO + 2 HCl --> CuCl2 + H2O;
0,0775---0,155 (mol)
nHCl=0,155(mol)=>mHCl=0,155*36,5=5,6575(g)
a. mdd X= 6,2+193,8= 200(g)=> C%X=6,2⋅100200=3,1%6,2⋅100200=3,1%
b. 2NaOH + CuSO4 ---> Na2SO4 + Cu(OH)2;
0,155--------0,0775------------0,0775----------0,0775 (mol)
Ta có: nCuSO4=200⋅16100⋅160=0,2(mol)200⋅16100⋅160=0,2(mol)
nNaOH=6,2400,155(mol)6,2400,155(mol)
Xét tỉ lệ:nNaOHnNaOHpt=0,1552<nCuSO4nCuSO4pt=0,21nNaOHnNaOHpt=0,1552<nCuSO4nCuSO4pt=0,21
=> CuSO4 dư. Sản phẩm tính theo NaOH.
=> nNa2SO4=0,155/2= 0,0775(mol)=> mNa2SO4=0,0775*142=11,005(g).
nCu(OH)2=0,155/2=0,0775(mol)=> mCu(OH)2=0,0775*98=7,595(g).
=> mdd sau pư= 200+200-7,595=392,405(g)
=> C%ddA=11,005⋅100392,405=2,8%
nHCl=0,6 mol
FeO+2HCl-->FeCl2+ H2O
x mol x mol
Fe2O3+6HCl-->2FeCl3+3H2O
x mol 2x mol
72x+160x=11,6 =>x=0,05 mol
A/ CFeCl2=0,05/0,3=1/6 M
CFeCl3=0,1/0,3=1/3 M
CHCl du=(0,6-0,4)/0,3=2/3 M
B/
NaOH+ HCl-->NaCl+H2O
0,2 0,2
2NaOH+FeCl2-->2NaCl+Fe(OH)2
0,1 0,05
3NaOH+FeCl3-->3NaCl+Fe(OH)3
0,3 0,1
nNaOH=0,6
CNaOH=0,6/1,5=0,4M
PTHH: \(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)
a) Ta có: \(n_{BaCl_2}=\frac{20,8}{208}=0,1\left(mol\right)\) \(\Rightarrow n_{H_2SO_4}=0,1mol\)
\(\Rightarrow m_{H_2SO_4}=98\cdot0,1=9,8\left(g\right)\) \(\Rightarrow m_{ddH_2SO_4}=\frac{9,8}{9,8\%}=100\left(g\right)\)
b) Theo PTHH: \(n_{BaCl_2}=n_{BaSO_4}=0,1mol\)
\(\Rightarrow m_{BaSO_4}=0,1\cdot233=23,3\left(g\right)\)
c) Theo PTHH: \(n_{BaCl_2}:n_{HCl}=1:2\) \(\Rightarrow n_{HCl}=0,2mol\)
\(\Rightarrow m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\)
Ta có: \(m_{dd}=m_{BaCl_2}+m_{ddH_2SO_4}-m_{BaSO_4}=20,8+100-23,3=97,5\left(g\right)\) \(\Rightarrow C\%_{HCl}=\frac{7,3}{97,5}\cdot100\approx7,49\%\)
nCuSO4= 32/160=0.2 mol
nNaOH= 2*0.25=0.5 mol
2NaOH + CuSO4 --> Na2SO4 + Cu(OH)2
Bđ: 0.5_______0.2
Pư: 0.4_______0.2_______0.2________0.2
Kt: 0.1________0________0.2________0.2
Cu(OH)2 -to-> CuO + H2O
0.2___________0.2
mCuO= 0.2*80=16g
mNaOH ( dư) = 0.1*40=4g
mNa2SO4= 0.2*142=28.4g
\(a.n_{NaOH}=\dfrac{20.20\%}{100\%.40}=0,1mol\\ 2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ n_{CuSO_4}=n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,05mol\\ m_{ddCuSO_4}=\dfrac{0,05.160}{10\%}\cdot100\%=80g\\ b.m_{Cu\left(OH\right)_2}=0,05.98=4,9g\\ c.C_{\%Na_2SO_4}=\dfrac{0,05.142}{20+80-4,9}\cdot100\%=7,46\%\)