balance so2+h2s-----s+h2o
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Answer:
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Answer:
SO₂ (g) + 2H₂S (g) ⟶ 2H₂O (l) + 3S (s)
Explanation:
SO₂ + H₂S ⟶ H₂O + S
In this reaction, sulphur dioxide ( SO₂ ) reacts with hydrogen sulphide ( H₂S ) to form water ( H₂O ) and sulphur ( S ).
Atoms in reactants side:
- S = 2
- O = 2
- H = 2
Atoms in products side:
- S = 1
- O = 1
- H = 2
By law of conservation of mass, we know that the number of atoms of each element in reactants side must be equal to number of atoms in products side.
In product side, 1 oxygen atom is missing. So to balance it, we multiply H₂O by 2.
SO₂ + H₂S ⟶ 2H₂O + S
Atoms in reactants side:
- S = 2
- O = 2
- H = 2
Atoms in products side:
- S = 1
- O = 2
- H = 4
In reactant side, 2 hydrogen atoms are missing. So, to balance it, we multiply H₂S by 2.
SO₂ + 2H₂S ⟶ 2H₂O + S
In product side, 2 sulphur atoms are missing, so we multiply S by 3 to balance both sides.
SO₂ (g) + 2H₂S (g) ⟶ 2H₂O (l) + 3S (s)
This is the balanced chemical equation.