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