To analyze the chemical equation 2H₂S + 3O₂ → 2H₂O + 2SO₂, we need to count the number of atoms of each element on both the reactant and product sides.
On the left (reactant) side:
- For Hydrogen (H): 2 moles of H₂S contain 2 × 2 = 4 H atoms.
- For Sulfur (S): 2 moles of H₂S contain 2 S atoms.
- For Oxygen (O): 3 moles of O₂ contain 3 × 2 = 6 O atoms.
So, on the left side, we have:
- Hydrogen (H): 4
- Sulfur (S): 2
- Oxygen (O): 6
On the right (product) side:
- For Hydrogen (H): 2 moles of H₂O contain 2 × 2 = 4 H atoms.
- For Sulfur (S): 2 moles of SO₂ contain 2 S atoms.
- For Oxygen (O): 2 moles of H₂O contain 2 O atoms, and 2 moles of SO₂ contain 2 × 2 = 4 O atoms. So, total O atoms = 2 + 4 = 6 O atoms.
Therefore, on the right side, we have:
- Hydrogen (H): 4
- Sulfur (S): 2
- Oxygen (O): 6
In conclusion, both sides of the equation have:
- Hydrogen (H): 4
- Sulfur (S): 2
- Oxygen (O): 6
This confirms that the equation is balanced, with the same number of each type of atom present on both sides.