To obtain pure lead, lead (II) sulfide is burned in an atmosphere of pure oxygen. The products of the reaction are lead and sulfur trioxide $\left( SO _3\right)$. Write a balanced chemical equation for this process.

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kalleighfaith01 kalleighfaith01

03-11-2026
Chemistry

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To obtain pure lead, lead (II) sulfide is burned in an atmosphere of pure oxygen. The products of the reaction are lead and sulfur trioxide $\left( SO _3\right)$. Write a balanced chemical equation for this process.

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-03T14:09:41-05:00

The reaction components consist of lead(II) sulfide and oxygen as reactants, producing pure lead and sulfur trioxide.
Counting the atoms for each element on both sides of the provided equation $2PbS + 3O_2 \longrightarrow 2Pb + 2SO_3$ confirms that there are 2 Pb atoms, 2 S atoms, and 6 O atoms on both sides.
Because the atoms are conserved, the provided equation is already correctly balanced.
The final balanced chemical equation is $\boxed{2PbS + 3O_2 \longrightarrow 2Pb + 2SO_3}$.

Explanation

Analysis of the Chemical Reaction

The goal is to determine if the chemical equation provided for the production of pure lead from lead(II) sulfide is balanced, and if so, confirm its correctness. The reaction involves lead(II) sulfide ($PbS$) reacting with oxygen gas ($O_2$) to produce metallic lead ($Pb$) and sulfur trioxide ($SO_3$). The provided equation is:
\[2PbS + 3O_2 \longrightarrow 2Pb + 2SO_3\]

Verification of the Balanced Equation

To determine if an equation is balanced, we must ensure that the number of atoms for each element is identical on both sides of the reaction. We list the number of atoms for each element on the reactant side (left) and the product side (right):

Reactants ($2PbS + 3O_2$):

Lead ($Pb$): $2 \times 1 = 2$
Sulfur ($S$): $2 \times 1 = 2$
Oxygen ($O$): $3 \times 2 = 6$

Products ($2Pb + 2SO_3$):

Lead ($Pb$): $2 \times 1 = 2$
Sulfur ($S$): $2 \times 1 = 2$
Oxygen ($O$): $2 \times 3 = 6$

Comparing the counts, we see that for every element (Lead, Sulfur, and Oxygen), the number of atoms on the reactant side is equal to the number of atoms on the product side.

Conclusion

Since the number of atoms of lead, sulfur, and oxygen are identical on both sides of the equation, the equation is already perfectly balanced. The chemical reaction accurately represents the transformation where $2$ units of lead(II) sulfide react with $3$ units of oxygen gas to produce $2$ units of pure lead and $2$ units of sulfur trioxide. The balanced chemical equation is $\boxed{2PbS + 3O_2 \longrightarrow 2Pb + 2SO_3}$.

Examples

In industrial metallurgy, balancing chemical equations is essential for calculating the exact mass of raw materials required and the quantity of products generated. For instance, when roasting metal sulfide ores like galena ($PbS$) in a furnace, engineers must precisely measure the amount of oxygen supplied to ensure efficient conversion while minimizing waste and environmental impact, such as sulfur dioxide emissions. This process ensures that extraction is cost-effective and compliant with safety standards.