a) 2Al+ 3H2SO4-> Al2(SO4)3+3H2

b) 2Fe(OH)3+ 3H2SO4-> Fe2(SO4)3+ 6H2O

a) 2 \(\rightarrow\)

b) 2 \(\rightarrow\)

BaO+H₂O➡ Ba(OH)₂

nBaO=30.6/(137+16)=0.2mol

a) Khi cho quỳ tím vào dd thì quỳ tím hóa xanh.

b) BaO + H₂O ---> Ba(OH)₂

nBaO = nBa(OH)₂ = 30.6 / 153 = 0.2 mol.

C_M Ba(OH)_{2 =} 0.2 / 0.56 = 5/14 M.

c) Ba(OH)_{2 +} H₂SO₄ ---> BaSO₄ + 2H₂O

nBa(OH)_{2 =} nH₂SO₄ = 0.2 mol.

---> mH₂SO_{4 =} 0.2 x 98 = 19.6 g.

---> m dd H₂SO_{4 =} 19.6 x 100 / 39.2 = 50 g.

`Fe + 2HCl -> FeCl_2 + H_2`

`2Na + 2HCl -> 2NaCl + H_2`

`2Al + 6HCl -> 2AlCl_3 + 3H_2`

`2Al(OH)_3 + 3H_2SO_4 -> Al_2(SO_4)_3 + 6H_2O`

Fe + 2HCl → FeCl₂ + H₂
2Na + 2HCl → 2NaCl + H₂
2Al + 6HCl → 2AlCl₃ + 3H₂
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(2Na+2HCl\rightarrow2NaCl+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)

a) \(K_2O+H_2O->2KOH\)

b) \(2Fe+3Cl_2->2FeCl_3\)

c) \(2Al\left(OH\right)_3+3H_2SO_4->Al_2\left(SO_4\right)_3+6H_2O\)

d) \(4FeS_2+11O_2->2Fe_2O_3+8SO_2\)

e)\(2Al+6HCl->2AlCl_3+3H_2\)

f) \(2KClO_3->KCl+3O_2\)

g) \(4P+5O_2->2P_2O_5\)

h) \(3Fe+2O_2->Fe_3O_4\)

i) \(2Al+3CuSO_4->Al_2\left(SO_4\right)_3+3Cu\)

j) \(2K+2H_2O->2KOH+H_2\)

2Al + 6H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

(5x-2y) Al + (18x-6y) HNO₃ → (5x-2y) Al(NO₃)₂ + 3N_xO_y + (9x-3y) H₂O

2M + 2n H₂SO₄ → M₂(SO₄)_n + nSO₂ +2n H₂O

lưu huỳnh oxi đồng

canxi sắt

mặt trời đồng

1+1=3

hok tốt

hello Duy

hello Duy

hello Duy

$\rm u)S + 2HNO_3 \rightarrow H_2SO_4 + 2NO \uparrow$

$\rm v)3Ba(NO_3)_2 + Al_2(SO_4)_3 \rightarrow 3BaSO_4 \downarrow + 2Al(NO_3)_3$
$\rm w)3Ba(OH)_2 + P_2O_5 \rightarrow Ba_3(PO_4)_3 \downarrow + 3H_2O$
$\rm x)2Fe_2O_3 + 3C \xrightarrow{t^o} 4Fe + 3CO_2 \uparrow$

pt: 2Al + 3H₂SO₄ => Al₂(SO₄)₃ + 3H₂

nAl = \(\dfrac{5,4}{27}=0,2mol\)

a) Theo pt: nH₂ = \(\dfrac{3}{2}nAl=\dfrac{3}{2}.0,2=0,3mol\)

=> VH₂ = 0,3.22,4 = 6,72 lít

b) Theo pt : nAl₂(SO₄)₃ = \(\dfrac{1}{2}nAl=0,1mol\)

=> mAl₂SO₄ = 0,1.342 = 34,2 g

a) Fe₂O₃ + 3H₂ -----> 2Fe + 3H₂O

1 mol 3 mol 2 mol 3 mol

0.2 mol 0.4mol

n_Fe=22.4/56=0.4 mol

b)m Fe₂O3 =n.M=0.2.160=32(g)

c) Fe + H₂SO₄ ------>FeSO₄ +H₂

0.4 mol 0.4mol

500ml=0.5 lít

C_{M= n/V=0.4/0.5=0.8M}