1)

a, \(n_{Al}=\dfrac{15,3}{102}=0,15\left(mol\right)\)

PTHH: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

Mol:     0,15         0,9        0,3      

\(m_{ddHCl}=\dfrac{0,9.36,5.100}{20}=164,25\left(g\right)\)

b, m_{dd sau pứ} = 15,3 + 164,25 = 179,55 (g)

c, \(C\%_{ddAlCl_3}=\dfrac{0,3.133,5.100\%}{179,55}=22,31\%\)

2)

a, \(m_{HCl}=54,75.20\%=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: Al₂O₃  + 6HCl → 2AlCl₃ + 3H₂O

Mol:      0,05        0,3          0,1

\(m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)

b, m_{dd sau pứ} = 5,1 + 54,75 = 59,85 (g) 

\(C\%_{ddAlCl_3}=\dfrac{0,1.133,5.100\%}{59,85}=22,31\%\)

nAl2O3= 10,2/102= 0,1(mol)

a) PTHH: Al2O3 + 6 HCl -> 2 AlCl3 + 3 H2O

0,1_______0,6_______0,2_________0,3(mol)

mHCl=0,6.36,5= 21,9(g)

=>mddHCl= (21,9.100)/7,3=300(g)

b) mddsau= mAl2O3 + mddHCl= 10,2+300=310,2(g)

c) mAlCl3= 133,5.0,2=26,7(g)

=>C%ddAlCl3= (26,7/310,2).100=8,607%

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

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

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

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

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

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

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

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

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

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

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)

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

n H2SO4=\(\dfrac{10\%.490}{2+32+16.4}=0,5mol\)

n Al2O3 =\(\dfrac{10,2}{27.2+16.3}=0,1mol\)

       \(Al_2O_3+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2O\)

bđ    0,1............0,5

pư    0,1............0,3..................0,1

spu   0 ................0,2................0,1 

=> sau pư gồm H2SO4 dư , Al2(S04)3 và H2O

m H2SO4 dư = \(0,2.\left(2+32+16.3\right)=19,6g\)

m Al2(SO4)3 = \(0,1\left(27.2+32.3+16.4.3\right)=34,2g\)

m dd = \(490+10,2=500,2g\)

% Al2(SO4)3 = \(\dfrac{34,2}{500,2}.100\sim6,84\%\)

% H2SO4 dư = \(\dfrac{19,6}{500,2}.100\sim3,92\%\)

\(a.CaO+2HCl\rightarrow CaCl_2+H_2O\\ n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ n_{CaCl_2}=n_{CaO}=0,2\left(mol\right)\\ m_{CaCl_2}=111.0,2=22,2\left(g\right)\)

Tên muối: Canxi clorua

\(b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\)