PTHH: Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
ddA: ddFeCl3
a) nFe2O3= m/M=12/160=0,075 (mol)
Từ PTHH=> nHCl=6nFe2O3=6.0,075=0,45 (mol)
mHCl=n.M=0,45.36,5=16,425 (mol)
C%HCl= \(\dfrac{m_{HCl}}{m_{ddHCl}}.100\%\)= \(\dfrac{16,425}{400}.100\%\simeq4,11\%\)
b)PTHH: FeCl3 + 3NaOH --> 3NaCl + Fe(OH)3\(\downarrow\)
Chất rắn: Fe(OH)3
Từ PTHH (a)=> nFeCl3=2nFe2O3=2.0,075=0,15 (mol)
Từ PTHH (b)=> nFe(OH)3=nFeCl3=0,15 (mol)
=> nNaOH=3nFeCl3=3.0,15=0,45 (mol)
mFe(OH)3=n.M=0,15.107=16,05 (g)
Hay a=16,05 (g)
CMNaOH=\(\dfrac{n_{NaOH}}{V_{ddNaOH}}=\dfrac{0,45}{0,3}=1,5\left(M\right)\)
nFe2O3 = \(\dfrac{12}{160}\) = 0,075 mol
Fe2O3 +6 HCl -> 2FeCl3 + 3H2O
0,075-->0,45 ---> 0,15
a) C%HCl = \(\dfrac{0,45.36,5}{400}.100\%\) = 4,10625 %
b) dd A là FeCl3
FeCl3 +3 NaOH -> 3NaCl + Fe(OH)3 \(\downarrow\)
0,15-->0,45 ------------------>0,15 mol
=> a = 0,15 . 107 = 16,05 g
=> CMNaOH = \(\dfrac{0,45}{0,3}\) = 1,5 M
Fe2O3 + 6HCl → 2FeCl3 + 3H2O (1)
\(n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\)
a) Theo PT1: \(n_{HCl}=6n_{Fe_2O_3}=6\times0,075=0,45\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,45\times36,5=16,425\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{16,425}{400}\times100\%=4,10625\left(M\right)\)
b) FeCl3 + 3NaOH → 3NaCl + Fe(OH)3↓ (2)
Theo PT1: \(n_{FeCl_3}=2n_{Fe_2O_3}=2\times0,075=0,15\left(mol\right)\)
Ta có: \(n_{FeCl_3}\left(2\right)=n_{FeCl_3}\left(1\right)=0,15\left(mol\right)\)
Theo PT2: \(n_{NaOH}=3n_{FeCl_3}=3\times0,15=0,45\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,45}{0,3}=1,5\left(M\right)\)
Theo PT2: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe\left(OH\right)_3}=0,15\times107=16,05\left(g\right)\)
Vậy \(a=16,05\left(g\right)\)
