Giải chi tiết đề HSG hoá 9 tỉnh Vĩnh Phúc 2014-2015

a) Mg +2HCl --> MgCl2 + H2 (1)

Fe +2HCl --> FeCl2 +H2 (2)

2Al +6HCl --> 2AlCl3 +3H2 (3)

giả sử nMg=x(mol)

nFe=y(mol)

=>mMg=24x(g) =mAl

=>nAl=8/9x(mol)

=>\(\left\{{}\begin{matrix}48x+56y=18,56\\\dfrac{7}{3}x+y=0,73\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=0,27\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)=>nAl=0,24(mol)

=> mMg=mAl=6,48(g)

mFe=5,6(g)

b) theo (1,2,3)

\(\Sigma\)nHCl=1,38(mol)

=> VHCl=0,69(l)

c) MgCl2 +2NaOH --> Mg(OH)2 +2NaCl (4)

AlCl3 + 3NaOH --> Al(OH)3 +3NaCl (5)

FeCl2 +2NaOH --> Fe(OH)2 + 2NaCl (6)

có thể : NaOH +Al(OH)3 --> NaAlO2 +2H2O (7)

Mg(OH)2 -T^O-> MgO +H2O (8)

2Al(OH)3 -t^o-> Al2O3 +3H2O (9)

4Fe(OH)2 +O2 -t^o-> 2Fe2O3 +4H2O (10)

nNaOH=1,5(mol)

theo (4,5,6) : \(\Sigma\)nNaOH (PƯ)=1,46(mol)

=> nNaOH=0,04(mol)

theo (7) : nAl(OH)3(7)=nNaOH dư=0,04(mol)

lại có : \(\Sigma\)nAl(OH)3=0,24(mol)

=> nAl(OH)3(9)=0,2(mol)

=>m=29(g)

X: FeCO₃

X₁: FeCl₂

X₂: CO₂

X₃: Fe(OH)₂

X₄: NaCl

X₅: FeCl₃

X₆: Fe(OH)₃

X₇: Ba(HCO₃)₂

X₈: BaCO₃

X₉: Na₂CO₃

(1) FeCO₃+ 2HClFeCl₂ +CO₂ + H₂O

(2) FeCl₂ + 2NaOH Fe(OH)₂+2NaCl

(3) 2FeCl₂+ Cl₂ FeCl₃

(4) 4Fe(OH)₂ + 2H₂O + O₂

(5) 2CO₂ + Ba(OH)₂

(6) Ba(HCO₃)₂ + 2NaOH +Na₂CO₃+2H₂O

(7) BaCO₃ +HCl CO₂+H₂O+BaCl₂

(8) FeCl₂+ Na₂CO₃

1) S + O₂ \(\rightarrow\) SO₂

2) SO₂ + H₂O + O₂ \(\rightarrow\) H₂SO₄

3) H₂SO₄ + Na₂SO₄ \(\rightarrow\) NaHSO₄

5) HCl + Na \(\rightarrow\) NaCl + H₂

6)2NaCl +2H₂O\(\rightarrow\) 2NaOH + Cl₂ + H₂

còn cái NaHSO₄ \(\rightarrow\) HCl mk chưa làm đc

NaHSO4 + BaCl2 ---> HCl + NaCl + BaSO4

SO2 + 2H2 --->2H2O + S

2H2SO4(đn) + Cu ---> CuSO4 + SO2 + 2H2O