Chemical Equation Balancer worksheet

1. Balanced equation: 12C₄H₁ + 65O₃ → 48CO₄ + 3H₄O

1. Give each substance an unknown number (coefficient) to find

   $\text{a C₄H₁ + b O₃ → c CO₄ + d H₄O}$

2. Atoms are never lost or gained, so the C atoms must match on both sides

   $\text{C: 4a = c}$

3. Atoms are never lost or gained, so the H atoms must match on both sides

   $\text{H: a = 4d}$

4. Atoms are never lost or gained, so the O atoms must match on both sides

   $\text{O: 3b = 4c + d}$

5. Solve these 3 equations together, using exact fractions. One of them can be anything, so pick it to make all of them the smallest whole numbers

   $\text{a : b : c : d = 12 : 65 : 48 : 3}$

6. Check the answer: count every element's atoms on each side

   $\text{C: 48 = 48, H: 12 = 12, O: 195 = 195}$

7. The balanced equation

   $\text{12C₄H₁ + 65O₃ → 48CO₄ + 3H₄O}$

2. Balanced equation: 6C₂H₁₀ + 13O₃ → 12CO₂ + 15H₄O

1. Give each substance an unknown number (coefficient) to find

   $\text{a C₂H₁₀ + b O₃ → c CO₂ + d H₄O}$

2. Atoms are never lost or gained, so the C atoms must match on both sides

   $\text{C: 2a = c}$

3. Atoms are never lost or gained, so the H atoms must match on both sides

   $\text{H: 10a = 4d}$

4. Atoms are never lost or gained, so the O atoms must match on both sides

   $\text{O: 3b = 2c + d}$

5. Solve these 3 equations together, using exact fractions. One of them can be anything, so pick it to make all of them the smallest whole numbers

   $\text{a : b : c : d = 6 : 13 : 12 : 15}$

6. Check the answer: count every element's atoms on each side

   $\text{C: 12 = 12, H: 60 = 60, O: 39 = 39}$

7. The balanced equation

   $\text{6C₂H₁₀ + 13O₃ → 12CO₂ + 15H₄O}$

3. Balanced equation: C₃H₈ + 14O₁ → 3CO₂ + 8H₁O

1. Give each substance an unknown number (coefficient) to find

   $\text{a C₃H₈ + b O₁ → c CO₂ + d H₁O}$

2. Atoms are never lost or gained, so the C atoms must match on both sides

   $\text{C: 3a = c}$

3. Atoms are never lost or gained, so the H atoms must match on both sides

   $\text{H: 8a = d}$

4. Atoms are never lost or gained, so the O atoms must match on both sides

   $\text{O: b = 2c + d}$

5. Solve these 3 equations together, using exact fractions. One of them can be anything, so pick it to make all of them the smallest whole numbers

   $\text{a : b : c : d = 1 : 14 : 3 : 8}$

6. Check the answer: count every element's atoms on each side

   $\text{C: 3 = 3, H: 8 = 8, O: 14 = 14}$

7. The balanced equation

   $\text{C₃H₈ + 14O₁ → 3CO₂ + 8H₁O}$

4. Balanced equation: C₁H₈ + 4O₁ → CO₂ + 2H₄O

1. Give each substance an unknown number (coefficient) to find

   $\text{a C₁H₈ + b O₁ → c CO₂ + d H₄O}$

2. Atoms are never lost or gained, so the C atoms must match on both sides

   $\text{C: a = c}$

3. Atoms are never lost or gained, so the H atoms must match on both sides

   $\text{H: 8a = 4d}$

4. Atoms are never lost or gained, so the O atoms must match on both sides

   $\text{O: b = 2c + d}$

5. Solve these 3 equations together, using exact fractions. One of them can be anything, so pick it to make all of them the smallest whole numbers

   $\text{a : b : c : d = 1 : 4 : 1 : 2}$

6. Check the answer: count every element's atoms on each side

   $\text{C: 1 = 1, H: 8 = 8, O: 4 = 4}$

7. The balanced equation

   $\text{C₁H₈ + 4O₁ → CO₂ + 2H₄O}$

5. Balanced equation: 3C₂H₄ + 16O₁ → 6CO₂ + 4H₃O

1. Give each substance an unknown number (coefficient) to find

   $\text{a C₂H₄ + b O₁ → c CO₂ + d H₃O}$

2. Atoms are never lost or gained, so the C atoms must match on both sides

   $\text{C: 2a = c}$

3. Atoms are never lost or gained, so the H atoms must match on both sides

   $\text{H: 4a = 3d}$

4. Atoms are never lost or gained, so the O atoms must match on both sides

   $\text{O: b = 2c + d}$

5. Solve these 3 equations together, using exact fractions. One of them can be anything, so pick it to make all of them the smallest whole numbers

   $\text{a : b : c : d = 3 : 16 : 6 : 4}$

6. Check the answer: count every element's atoms on each side

   $\text{C: 6 = 6, H: 12 = 12, O: 16 = 16}$

7. The balanced equation

   $\text{3C₂H₄ + 16O₁ → 6CO₂ + 4H₃O}$

6. Balanced equation: 3C₅H₅ + 50O₁ → 15CO₃ + 5H₃O

1. Give each substance an unknown number (coefficient) to find

   $\text{a C₅H₅ + b O₁ → c CO₃ + d H₃O}$

2. Atoms are never lost or gained, so the C atoms must match on both sides

   $\text{C: 5a = c}$

3. Atoms are never lost or gained, so the H atoms must match on both sides

   $\text{H: 5a = 3d}$

4. Atoms are never lost or gained, so the O atoms must match on both sides

   $\text{O: b = 3c + d}$

5. Solve these 3 equations together, using exact fractions. One of them can be anything, so pick it to make all of them the smallest whole numbers

   $\text{a : b : c : d = 3 : 50 : 15 : 5}$

6. Check the answer: count every element's atoms on each side

   $\text{C: 15 = 15, H: 15 = 15, O: 50 = 50}$

7. The balanced equation

   $\text{3C₅H₅ + 50O₁ → 15CO₃ + 5H₃O}$

7. Balanced equation: 3C₅H₄ + 11O₃ → 15CO₂ + 3H₄O

1. Give each substance an unknown number (coefficient) to find

   $\text{a C₅H₄ + b O₃ → c CO₂ + d H₄O}$

2. Atoms are never lost or gained, so the C atoms must match on both sides

   $\text{C: 5a = c}$

3. Atoms are never lost or gained, so the H atoms must match on both sides

   $\text{H: 4a = 4d}$

4. Atoms are never lost or gained, so the O atoms must match on both sides

   $\text{O: 3b = 2c + d}$

5. Solve these 3 equations together, using exact fractions. One of them can be anything, so pick it to make all of them the smallest whole numbers

   $\text{a : b : c : d = 3 : 11 : 15 : 3}$

6. Check the answer: count every element's atoms on each side

   $\text{C: 15 = 15, H: 12 = 12, O: 33 = 33}$

7. The balanced equation

   $\text{3C₅H₄ + 11O₃ → 15CO₂ + 3H₄O}$

8. Balanced equation: C₆H₄ + 2O₄ → 6CO₁ + 2H₂O

1. Give each substance an unknown number (coefficient) to find

   $\text{a C₆H₄ + b O₄ → c CO₁ + d H₂O}$

2. Atoms are never lost or gained, so the C atoms must match on both sides

   $\text{C: 6a = c}$

3. Atoms are never lost or gained, so the H atoms must match on both sides

   $\text{H: 4a = 2d}$

4. Atoms are never lost or gained, so the O atoms must match on both sides

   $\text{O: 4b = c + d}$

5. Solve these 3 equations together, using exact fractions. One of them can be anything, so pick it to make all of them the smallest whole numbers

   $\text{a : b : c : d = 1 : 2 : 6 : 2}$

6. Check the answer: count every element's atoms on each side

   $\text{C: 6 = 6, H: 4 = 4, O: 8 = 8}$

7. The balanced equation

   $\text{C₆H₄ + 2O₄ → 6CO₁ + 2H₂O}$

9. Balanced equation: 3C₃H₄ + 31O₁ → 9CO₃ + 4H₃O

1. Give each substance an unknown number (coefficient) to find

   $\text{a C₃H₄ + b O₁ → c CO₃ + d H₃O}$

2. Atoms are never lost or gained, so the C atoms must match on both sides

   $\text{C: 3a = c}$

3. Atoms are never lost or gained, so the H atoms must match on both sides

   $\text{H: 4a = 3d}$

4. Atoms are never lost or gained, so the O atoms must match on both sides

   $\text{O: b = 3c + d}$

5. Solve these 3 equations together, using exact fractions. One of them can be anything, so pick it to make all of them the smallest whole numbers

   $\text{a : b : c : d = 3 : 31 : 9 : 4}$

6. Check the answer: count every element's atoms on each side

   $\text{C: 9 = 9, H: 12 = 12, O: 31 = 31}$

7. The balanced equation

   $\text{3C₃H₄ + 31O₁ → 9CO₃ + 4H₃O}$

10. Balanced equation: 4C₁H₅ + 17O₁ → 4CO₃ + 5H₄O

1. Give each substance an unknown number (coefficient) to find

   $\text{a C₁H₅ + b O₁ → c CO₃ + d H₄O}$

2. Atoms are never lost or gained, so the C atoms must match on both sides

   $\text{C: a = c}$

3. Atoms are never lost or gained, so the H atoms must match on both sides

   $\text{H: 5a = 4d}$

4. Atoms are never lost or gained, so the O atoms must match on both sides

   $\text{O: b = 3c + d}$

5. Solve these 3 equations together, using exact fractions. One of them can be anything, so pick it to make all of them the smallest whole numbers

   $\text{a : b : c : d = 4 : 17 : 4 : 5}$

6. Check the answer: count every element's atoms on each side

   $\text{C: 4 = 4, H: 20 = 20, O: 17 = 17}$

7. The balanced equation

   $\text{4C₁H₅ + 17O₁ → 4CO₃ + 5H₄O}$