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PTHH: \(3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\)
Ta có: \(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)
\(\Rightarrow n_{FeCl_3}=n_{Fe\left(OH\right)_3}=\dfrac{1}{12}\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_3}=\dfrac{1}{12}\cdot162,5\approx13,54\left(g\right)\\m_{Fe\left(OH\right)_3}=\dfrac{1}{12}\cdot107\approx8,92\left(g\right)\end{matrix}\right.\)
nNaOH = m/M = 10/(23 +16 + 1) = 0,25 (mol)
Ta có PTHH: 3NaOH + FeCl3 ------> Fe(OH)3 + 3NaCl
Theo PT: 3 - 1 - 1 (mol)
BC: 0.25 - 0.083 - 0.083 (mol)
Suy ra: mFeCl3 = n x M = 0.083 x (56 + 35,5 x 3) = 13,4875 (g)
mFe(OH)3 = n x M = 0,083 x (56+17 x 3) = 8,881 (g)
số mol NaOH là:\(n_{NaOH}=\frac{10}{23+16+1}=0,25\left(mol\right)\)
PTHH\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(m_{FeCl_3}=n.M=\frac{0.25}{3}\cdot\left(56+35,5\cdot3\right)\approx13,54\left(g\right)\)
\(m_{Fe\left(OH\right)_3}=n.M=\frac{0.25}{3}\cdot\left(56+\left(16+1\right)\cdot3\right)\approx8,91\left(g\right)\)
\(m_{NaCl}=n.M=0.25\cdot\left(23+35.5\right)=14.625\left(g\right)\)
\(n_{Fe_2O_3}=\dfrac{3.2}{160}=0.02\left(mol\right)\)
\(n_{HCl}=\dfrac{2.19}{36.5}=0.06\left(mol\right)\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(1...........6\)
\(0.02...........0.06\)
Lập tỉ lệ : \(\dfrac{0.02}{1}>\dfrac{0.06}{6}\Rightarrow Fe_2O_3dư\)
\(n_{Fe_2O_3\left(dư\right)}=0.02-\dfrac{0.06}{6}=0.01\left(mol\right)\)
\(m_{Fe_2O_3\left(dư\right)}=0.01\cdot160=1.6\left(g\right)\)
\(m_{FeCl_3}=0.02\cdot162.5=3.25\left(g\right)\)
\(m_{H_2O}=0.03\cdot18=0.54\left(g\right)\)
\(n_{NaOH}=\dfrac{m}{M}=\dfrac{4}{40}=0,1mol\)
\(n_{H_2SO_4}=\dfrac{9,8}{98}=0,1mol\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(\dfrac{0,1}{2}\) < \(\dfrac{0,1}{1}\) ( mol )
0,1 0,05 ( mol )
Chất còn dư là H2SO4
Chất phản ứng hết là NaOH
\(m_{Na_2SO_4}=n.M=0,05.142=7,1g\)
nNaOH = 4/40 = 0,1 (mol)
nH2SO4 = 9,8/98 = 0,1 (mol)
PTHH: 2NaOH + H2SO4 -> Na2SO4 + H2O
LTL: 0,1/2 < 0,1 => H2SO4 dư
nNa2SO4 = 0,1/2 = 0,05 (mol)
mNa2SO4 = 0,05 . 142 = 7,1 (g)
nFe=11,2/56=0,2 mol
nH2SO4=24,5/98=0,25
PTPƯ: Fe + H2SO4 ---> FeSO4 + H2
0,2 mol ----> 0,2 mol --------------------> 0,2 mol
Ta có Fe:H2SO4=0,2/1<0,25/1 (nên H2SO4 dư)
a, mH2SO4=(0,25-0,2).98=4,9 g
b, VH2=0,2.22,4=4,48 l
PTHH: \(Fe+H_2SO_{4\left(l\right)}\rightarrow FeSO_4+H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Sắt còn dư, Axit p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,25\left(mol\right)\\n_{Fe\left(dư\right)}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,25\cdot22,4=5,6\left(l\right)\\m_{Fe\left(dư\right)}=0,15\cdot56=8,4\left(g\right)\end{matrix}\right.\)
a)
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)
\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow3Na_2SO_4+2Al\left(OH\right)_3\downarrow\)
\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)
\(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
b) B gồm Fe(OH)2, Cu(OH)2
C gồm CuO, Fe2O3
a)\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)
\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow2Al\left(OH\right)_3+3Na_2SO_4\)
\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
\(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)
b)B là các chất sau: \(Fe\left(OH\right)_2;Cu\left(OH\right)_2;Al\left(OH\right)_3\)
C là các chất sau: \(Fe_2O_3;CuO;Al_2O_3\)
a) Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
\(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02\left(mol\right)\)
\(n_{HCl}=\dfrac{2,19}{36,5}=0,06\left(mol\right)\)
Xét tỉ lệ \(\dfrac{0,02}{1}>\dfrac{0,06}{6}\) => Fe2O3 dư, HCl hết
PTHH: Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
0,01<--0,06------->0,02---->0,03
=> \(m_{Fe_2O_3\left(dư\right)}=\left(0,02-0,01\right).160=1,6\left(g\right)\)
b) \(m_{FeCl_3}=0,02.162,5=3,25\left(g\right)\)
\(m_{H_2O}=0,03.18=0,54\left(g\right)\)
`Fe + H_2 SO_4 -> FeSO_4 + H_2`
`0,25` `0,25` `0,25` `(mol)`
`a)n_[Fe]=[22,4]/56=0,4(mol)`
`n_[H_2 SO_4]=[24,5]/98=0,25(mol)`
Có: `[0,4]/1 > [0,25]/1=>Fe` hết, `H_2 SO_4`
`=>m_[Fe(dư)]=(0,4-0,25).56=8,4(g)`
`b)V_[H_2]=0,25.22,4=5,6(l)`
Ko được ghi `Fe+H_2 SO_4->Fe_2 (SO_4)_3+H_2` vì đây là `H_2 SO_4` loãng
a) nNaOH= 6/40=0,15(mol)
nFeCl3=32,5/162,5= 0,2(mol)
PTHH: 3 NaOH + FeCl3 -> Fe(OH)3 + 3 NaCl
0,15________0,05____0,05________0,15(mol)
Ta có: 0,2/1 > 0,15/3
=> NaOH hết, FeCl3 dư
=> nFeCl3(dư)= 0,2-0,05=0,15(mol)
=> mFeCl3= 162,5.0,15=24,375(g)
b)m(kết tủa)= mFe(OH)3= 0,05.107= 5,35(g)
cảm ơn nhiều ạ