`Fe + 2HCl -> FeCl_2 + H_2`

`0,1`   `0,2`               `0,1`     `0,1`          `(mol)`

`n_[Fe]=[5,6]/56=0,1(mol)`

`a)m_[FeCl_2]=0,1.127=12,7(g)`

`b)C%_[HCl]=[0,2.36,5]/250 . 100=2,92%`

`c)C%_[FeCl_2]=[12,7]/[5,6+250-0,1.2].100~~4,97%`

\(n_{FeO}=\dfrac{10.8}{72}=0.15\left(mol\right)\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

\(0.15.......0.3.............0.15\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

\(C\%HCl=\dfrac{10.95}{100}\cdot100\%=10.95\%\)

\(m_{dd}=10.8+100=110.8\left(g\right)\)

\(m_{FeCl_2}=0.15\cdot127=19.05\left(g\right)\)

\(C\%FeCl_2=\dfrac{19.05}{110.8}\cdot100\%=17.19\%\)

C%HCl=10.95100⋅100%=10.95%C%HCl=10.95100⋅100%=10.95%

mdd=10.8+100=110.8(g)mdd=10.8+100=110.8(g)

mFeCl2=0.15⋅127=19.05(g)mFeCl2=0.15⋅127=19.05(g)

\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

______0,05------>0,15--------->0,05

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)

\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)

PTHH: Fe₂(SO₄)₃ + 6NaOH --> 2Fe(OH)₃\(\downarrow\) + 3Na₂SO₄

________0,05----------------------->0,1

=> mFe(OH)₃ = 0,1.107=10,7(g)

a) $2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
b)

n Al = 2,7/27 = 0,1(mol)

Theo PTHH :

n H2SO4 = 3/2 n Al = 0,15(mol)

=> C% H2SO4 = 0,15.98/441  .100% = 3,33%

c)

Theo PTHH :

n H2 = 3/2 n Al = 0,15(mol)

=> V H2 = 0,15.22,4 = 3,36(lít)

Ta có: \(n_{Fe_2O_3}=\dfrac{9,6}{160}=0,06\left(mol\right)\)

a. PTHH: Fe₂O₃ + 3H₂SO₄ ---> Fe₂(SO₄)₃ + 3H₂O (1)

Theo PT₍₁₎: \(n_{H_2SO_4}=3.n_{Fe_2O_3}=3.0,06=0,18\left(mol\right)\)

=> \(m_{H_2SO_4}=0,18.98=17,64\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{17,64}{m_{dd_{H_2SO_4}}}.100\%=9,8\%\)

=> \(m_{dd_{H_2SO_4}}=180\left(g\right)\)

b. Ta có: \(m_{dd_{Fe_2\left(SO_4\right)_3}}=9,6+180=189,6\left(g\right)\)

Theo PT₍₁₎: \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,06\left(mol\right)\)

=> \(m_{Fe_2\left(SO_4\right)_3}=0,06.400=24\left(g\right)\)

=> \(C_{\%_{Fe_2\left(SO_4\right)_3}}=\dfrac{24}{189,6}.100\%=12,66\%\)

c. PTHH: Fe₂(SO₄)₃ + 3BaCl₂ ---> 3BaSO₄↓ + 2FeCl₃ (2)

Theo PT₍₂₎: \(n_{BaSO_4}=3.n_{Fe_2\left(SO_4\right)_3}=3.0,06=0,18\left(mol\right)\)

=> \(m_{BaSO_4}=0,18.233=41,94\left(g\right)\)

Theo PT₍₂₎: \(n_{BaCl_2}=n_{BaSO_4}=0,18\left(mol\right)\)

=> \(m_{BaCl_2}=0,18.208=37,44\left(g\right)\)

Ta có: \(C_{\%_{BaCl_2}}=\dfrac{37,44}{m_{dd_{BaCl_2}}}.100\%=10,4\%\)

=> \(m_{dd_{BaCl_2}}=360\left(g\right)\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a.Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{ddHCl}=\dfrac{0,4.36,5.100}{7,3}=200\left(g\right)\\ c.m_{ddsau}=4,8+200-0,2.2=204,4\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,2.95}{204,4}.100\approx9,295\%\\ d.V_{ddHCl}=\dfrac{200}{1,05}=\dfrac{4000}{21}\left(ml\right)=\dfrac{4}{21}\left(l\right)\\ C_{MddHCl}=\dfrac{0,4}{\dfrac{4}{21}}=2,1\left(M\right)\)

Cho 10,8 g FeO t/d vừa đủ với 100g dung dịch axit clohidric
a. Tính khối lượng axít đã dùng, từ đó suy ra nồng độ % của dung dịch axit?
b. Tính nồng độ % của dung dịch muối thu được sau phản ứng?
Giải:
a) nFeo=10,8/72=0,15
Phương trình hóa học:FeO + 2HCl----->FeCl2+H2O
Theo Phương trình: 1mol 2 mol 1mol
Theo đề bài: 0,15mol 0,3mol 0,15mol
\Rightarrowmhcl=0,3*36,5=10,95g
\Rightarrow C%HCl=mHCl/mDung dich=10,95/100*100=10,95%
b)Từ câu a\Rightarrow mFeCl2=0,15*127=19,05
mdung dich=10,8+100=110,8g
\RightarrowC%FeCl2=19,05/110,8*100=17,19%

PTHH: FeO + 2HCl \(\rightarrow\) FeCl₂ + H₂O

Ta có: n _FeO =  \(\frac{10,8}{72}\) = 0,15 mol

Theo p.trình: n _HCl=2.n_FeO = 2.0,15=0,3 mol

\(\Rightarrow\) m_HCl = 0,3 . 36,5 = 10,95g

\(\Rightarrow\) C% _{dd HCl} = \(\frac{10,95}{100}.100\) = 10,95 %

Theo p.trình: n_FeCl2 = n_FeO = 0,15 mol

\(\Rightarrow\) m_FeCl2 = 0,15 . 127 = 19,05 (g)

m_{dd FeCl2}= 10,8 + 100= 110,8 g

\(\Rightarrow\) C% _{dd muối} = \(\frac{19,05}{110,8}.100\) = 17,19%

a. Ta có n_FeO = \(\dfrac{10,8}{72}\) = 0,15 ( mol )

FeO + 2HCl → FeCl₂ + H₂O

0,15 → 0,3 → 0,15 → 0,15

=> m_HCl = 0,3 . 36,5 = 10,95 ( gam )

=> C%_HCl = \(\dfrac{10,95}{100}\) . 100 = 10,95%

b, => m_FeCl2 = 0,15 . 127 = 19,05 ( gam )

Ta có M_{dung dịch} = M_{chất tham gia}

= 10,8 + 100

= 110,8 ( gam )

=> C%_HCl = \(\dfrac{19,05}{110,8}\) . 100 \(\approx\) 17,19%

a) Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

b) \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH: Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

                0,1---->0,3------->0,1

=> m = 0,1.342 = 34,2 (g)

c) \(C\%_{dd.H_2SO_4}=\dfrac{0,3.98}{120}.100\%=24,5\%\)

Al2O3+3H2SO4->Al2(SO4)3+3H2O

0,1------0,3------0,1-------------------0,3

n Al2O3=0,1 mol

m Al2(SO4)3=0,1.342=34,2g

C%=\(\dfrac{0,3.98}{120}100=24,5\%\)

\(a.Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2 O\\ b.n_{Al_2O_3}=0,1\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,1\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\\ c.n_{H_2SO_4}=3n_{Al_2O_3}=0,3\left(mol\right)\\ C\%_{H_2SO_4}=\dfrac{0,3.98}{120}.100=24,5\%\)