How do you balance these chemical equations: P + O₂ → P₄O₁₀; Mg + O₂ → MgO; HgO → Hg + O₂; Al₂O₃ → Al + O₂; BaCl₂ + H₂SO₄ → BaSO₄ + HCl?

What balancing is checking

A chemical equation is balanced when each element has the same number of atoms on the reactant side and the product side. You are allowed to change only coefficients (numbers in front), not subscripts inside formulas.

Balancing $P + O_2 \rightarrow P_4O_{10}$

The product has $4$ phosphorus atoms and $10$ oxygen atoms.

• Match phosphorus: put $4$ in front of $P$.
• Match oxygen: $10$ oxygens means $5O_2$.

Balanced: $$4P + 5O_2 \rightarrow P_4O_{10}$$

Balancing $Mg + O_2 \rightarrow MgO$

Oxygen is diatomic as $O_2$, but $MgO$ has 1 oxygen per unit.

• Put $2$ in front of $MgO$ to make oxygen count even.
• Then put $2$ in front of $Mg$.

Balanced: $$2Mg + O_2 \rightarrow 2MgO$$

Balancing $HgO \rightarrow Hg + O_2$

Oxygen appears as $O_2$ on the product side.

• Put $2$ in front of $HgO$ so there are $2$ oxygen atoms total.
• Then put $2$ in front of $Hg$ to match mercury.

Balanced: $$2HgO \rightarrow 2Hg + O_2$$

Balancing $Al_2O_3 \rightarrow Al + O_2$

This one often needs a least common multiple for oxygen.

• Oxygen: $Al_2O_3$ has $3$ O atoms, and $O_2$ has $2$ O atoms. The LCM of $3$ and $2$ is $6$.
• Put $2$ in front of $Al_2O_3$ to get $6$ oxygens.
• Put $3$ in front of $O_2$ to get $6$ oxygens on the right.
• Now aluminum: left has $2 \times 2 = 4$ Al atoms, so put $4$ in front of $Al$.

Balanced: $$2Al_2O_3 \rightarrow 4Al + 3O_2$$

Balancing $BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + HCl$

Treat the sulfate $SO_4$ as a unit because it stays together.

• $Ba$ is already 1 to 1.
• $SO_4$ is already 1 to 1.
• Left has $2$ chlorines in $BaCl_2$, so put $2$ in front of $HCl$.
• That also gives $2$ hydrogens, matching the $2$ in $H_2SO_4$.

Balanced: $$BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2HCl$$

Quick self-check you can do every time

After you finish, count each element on both sides (including diatomic elements like $O_2$). If every count matches and coefficients share no common factor, you are done.