Answer
Balanced chemical equation:
$2H_2S(g) + 3O_2(g) --\gt 2SO_2(g) + 2H_2O(g)$
Work Step by Step
1. Identify the chemical formulas for each reactant and product:
Reactants: Dihydrogen sulfide: $H_2S(g)$. Oxygen: $O_2(g)$
Products: Sulfur dioxide gas : $SO_2(g)$. Water vapor: $H_2O(g)$
2. Write the unbalanced reaction:
$H_2S(g) + O_2(g) --\gt SO_2(g) + H_2O(g)$
3. Balance the number of oxygen atoms by putting a "$\frac{3}{2}$" in front of $O_2$.
$H_2S(g) + \frac{3}{2} O_2(g) --\gt SO_2(g) + H_2O(g)$
4. To remove the fraction, multiply all coefficients by "2":
$2H_2S(g) + 3O_2(g) --\gt 2SO_2(g) + 2H_2O(g)$