Accepted answer
Answer included
Problem
A balancing equations worksheet tests whether each chemical equation is written with coefficients that conserve atoms (and therefore mass) in a closed system. Balance each equation below using the smallest whole-number stoichiometric coefficients.
Checkpoint for every line: after balancing, the number of atoms of each element on the reactant side must
equal the number of atoms of that element on the product side.
Visualization: atom counts before and after balancing
Method used on a balancing equations worksheet
- List each distinct element appearing in the reaction.
- Count atoms of each element on the reactant side and on the product side.
- Adjust coefficients only (numbers in front of formulas), never subscripts in the formulas.
- Start with elements that appear in the fewest compounds; leave hydrogen and oxygen for later in many cases.
- Clear fractions by multiplying all coefficients by the smallest common factor, then reduce if possible.
- Recheck all atom counts to confirm equality on both sides.
Worked example (system-of-equations view)
Balance Al + O2 → Al2O3 by assigning unknown coefficients:
\[ a \cdot \mathrm{Al} + b \cdot \mathrm{O_2} \rightarrow c \cdot \mathrm{Al_2O_3} \]
Atom-balance equations:
- Al: \(a = 2 \cdot c\)
- O: \(2 \cdot b = 3 \cdot c\)
Choose the smallest integer \(c\) that makes both equations integers. Setting \(c = 2\) gives: \(a = 2 \cdot 2 = 4\) and \(2 \cdot b = 3 \cdot 2 = 6\), so \(b = 3\).
Balanced result: 4 Al + 3 O2 → 2 Al2O3
Worksheet: practice equations
Balance each equation using the smallest whole-number coefficients.
| # | Unbalanced equation | Reaction type (hint) |
|---|---|---|
| 1 | H2 + O2 → H2O | Synthesis |
| 2 | Al + O2 → Al2O3 | Synthesis |
| 3 | KClO3 → KCl + O2 | Decomposition |
| 4 | C3H8 + O2 → CO2 + H2O | Combustion |
| 5 | Zn + HCl → ZnCl2 + H2 | Single replacement |
| 6 | BaCl2 + Na2SO4 → BaSO4 + NaCl | Double replacement |
| 7 | Na3PO4 + MgCl2 → NaCl + Mg3(PO4)2 | Double replacement |
| 8 | NH3 + O2 → NO + H2O | Redox / combustion-like |
Answer key (balanced equations)
| # | Balanced equation | Quick atom-check notes |
|---|---|---|
| 1 | 2 H2 + O2 → 2 H2O | H: 4 vs 4; O: 2 vs 2 |
| 2 | 4 Al + 3 O2 → 2 Al2O3 | Al: 4 vs 4; O: 6 vs 6 |
| 3 | 2 KClO3 → 2 KCl + 3 O2 | K: 2 vs 2; Cl: 2 vs 2; O: 6 vs 6 |
| 4 | C3H8 + 5 O2 → 3 CO2 + 4 H2O | C: 3 vs 3; H: 8 vs 8; O: 10 vs 10 |
| 5 | Zn + 2 HCl → ZnCl2 + H2 | Zn: 1 vs 1; H: 2 vs 2; Cl: 2 vs 2 |
| 6 | BaCl2 + Na2SO4 → BaSO4 + 2 NaCl | Ba: 1 vs 1; (SO4): 1 vs 1; Na: 2 vs 2; Cl: 2 vs 2 |
| 7 | 2 Na3PO4 + 3 MgCl2 → 6 NaCl + Mg3(PO4)2 | Na: 6 vs 6; P: 2 vs 2; O: 8 vs 8; Mg: 3 vs 3; Cl: 6 vs 6 |
| 8 | 4 NH3 + 5 O2 → 4 NO + 6 H2O | N: 4 vs 4; H: 12 vs 12; O: 10 vs 10 |
Common mistakes to avoid
- Changing subscripts (e.g., turning O2 into O3) changes the substance and is not allowed.
- Stopping after balancing one element; every element must match on both sides.
- Leaving fractional coefficients; multiply all coefficients by a common factor to make integers, then reduce if possible.
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