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Step-by-step solution for: Balancing Equations Practice online exercise for
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Show Answer Key & Explanations
Step-by-step solution for: Balancing Equations Practice online exercise for
To balance the chemical equations, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Here are the balanced equations with explanations:
---
- Reactants: Fe, H₂SO₄
- Products: Fe₂(SO₄)₃, H₂
Balancing:
- Start with Fe: There are 2 Fe atoms in Fe₂(SO₄)₃, so we need 2 Fe atoms on the left.
- Next, balance H and S: Each H₂SO₄ provides 2 H atoms and 1 S atom. Since there are 3 SO₄ groups in Fe₂(SO₄)₃, we need 3 H₂SO₄ molecules.
- Finally, balance H₂: From 3 H₂SO₄, we get 6 H atoms, which means we need 3 H₂ molecules.
Balanced Equation:
$$ \boxed{2 \text{Fe} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3 \text{H}_2} $$
---
- Reactants: C₂H₆, O₂
- Products: H₂O, CO₂
Balancing:
- Start with C: There are 2 C atoms in C₂H₆, so we need 2 CO₂ molecules.
- Next, balance H: There are 6 H atoms in C₂H₆, so we need 3 H₂O molecules.
- Finally, balance O: Each CO₂ has 2 O atoms (total 4 O atoms), and each H₂O has 1 O atom (total 3 O atoms), making a total of 7 O atoms needed. Since O₂ provides 2 O atoms per molecule, we need 7/2 = 3.5 O₂ molecules. To avoid fractions, multiply everything by 2.
Balanced Equation:
$$ \boxed{2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 6 \text{H}_2\text{O} + 4 \text{CO}_2} $$
---
- Reactants: KOH, H₃PO₄
- Products: K₃PO₄, H₂O
Balancing:
- Start with K: There are 3 K atoms in K₃PO₄, so we need 3 KOH molecules.
- Next, balance P: There is 1 P atom in H₃PO₄, so we need 1 H₃PO₄ molecule.
- Balance H and O: From 3 KOH, we get 3 H atoms, and from H₃PO₄, we get 3 more H atoms, totaling 6 H atoms, which means we need 3 H₂O molecules.
Balanced Equation:
$$ \boxed{3 \text{KOH} + \text{H}_3\text{PO}_4 \rightarrow \text{K}_3\text{PO}_4 + 3 \text{H}_2\text{O}} $$
---
- Reactants: SnO₂, H₂
- Products: Sn, H₂O
Balancing:
- Start with Sn: There is 1 Sn atom in SnO₂, so we need 1 Sn atom on the right.
- Next, balance O: There are 2 O atoms in SnO₂, so we need 2 H₂O molecules.
- Finally, balance H: Each H₂O has 2 H atoms, so we need 2 H₂ molecules.
Balanced Equation:
$$ \boxed{\text{SnO}_2 + 2 \text{H}_2 \rightarrow \text{Sn} + 2 \text{H}_2\text{O}} $$
---
- Reactants: NH₃, O₂
- Products: NO, H₂O
Balancing:
- Start with N: There is 1 N atom in NH₃, so we need 1 NO molecule.
- Next, balance H: There are 3 H atoms in NH₃, so we need 3/2 H₂O molecules. To avoid fractions, multiply everything by 2.
- Balance O: From 2 NO, we have 2 O atoms, and from 3 H₂O, we have 3 O atoms, totaling 5 O atoms. Since O₂ provides 2 O atoms per molecule, we need 5/2 = 2.5 O₂ molecules. Multiply everything by 2 again to eliminate the fraction.
Balanced Equation:
$$ \boxed{4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O}} $$
---
- Reactants: KNO₃, H₂CO₃
- Products: K₂CO₃, HNO₃
Balancing:
- Start with K: There are 2 K atoms in K₂CO₃, so we need 2 KNO₃ molecules.
- Next, balance C: There is 1 C atom in H₂CO₃, so we need 1 H₂CO₃ molecule.
- Balance H and O: From 2 KNO₃, we get 2 HNO₃ molecules, and from H₂CO₃, we get 1 CO₃ group, which matches K₂CO₃.
Balanced Equation:
$$ \boxed{2 \text{KNO}_3 + \text{H}_2\text{CO}_3 \rightarrow \text{K}_2\text{CO}_3 + 2 \text{HNO}_3} $$
---
- Reactants: B₂Br₆, HNO₃
- Products: B(NO₃)₃, HBr
Balancing:
- Start with B: There are 2 B atoms in B₂Br₆, so we need 2 B(NO₃)₃ molecules.
- Next, balance Br: There are 6 Br atoms in B₂Br₆, so we need 6 HBr molecules.
- Balance N and O: Each B(NO₃)₃ has 3 NO₃ groups, so we need 6 HNO₃ molecules.
Balanced Equation:
$$ \boxed{\text{B}_2\text{Br}_6 + 6 \text{HNO}_3 \rightarrow 2 \text{B(NO}_3)_3 + 6 \text{HBr}} $$
---
- Reactants: BF₃, Li₂SO₃
- Products: B₂(SO₃)₃, LiF
Balancing:
- Start with B: There are 2 B atoms in B₂(SO₃)₃, so we need 2 BF₃ molecules.
- Next, balance F: Each BF₃ has 3 F atoms, so we need 6 LiF molecules.
- Balance Li and S: Each Li₂SO₃ has 1 S atom, so we need 3 Li₂SO₃ molecules to provide 3 SO₃ groups for B₂(SO₃)₃. This also gives us 6 Li atoms, matching the 6 LiF molecules.
Balanced Equation:
$$ \boxed{2 \text{BF}_3 + 3 \text{Li}_2\text{SO}_3 \rightarrow \text{B}_2(\text{SO}_3)_3 + 6 \text{LiF}} $$
---
- Reactants: (NH₄)₃PO₄, Pb(NO₃)₄
- Products: Pb₃(PO₄)₄, NH₄NO₃
Balancing:
- Start with Pb: There are 3 Pb atoms in Pb₃(PO₄)₄, so we need 3 Pb(NO₃)₄ molecules.
- Next, balance P: There are 4 P atoms in Pb₃(PO₄)₄, so we need 4 (NH₄)₃PO₄ molecules.
- Balance N, H, and O: Each (NH₄)₃PO₄ provides 3 NH₄ groups, so we need 12 NH₄NO₃ molecules. Each Pb(NO₃)₄ provides 4 NO₃ groups, so we get 12 NO₃ groups in total, matching the 12 NH₄NO₃ molecules.
Balanced Equation:
$$ \boxed{4 (\text{NH}_4)_3\text{PO}_4 + 3 \text{Pb(NO}_3)_4 \rightarrow \text{Pb}_3(\text{PO}_4)_4 + 12 \text{NH}_4\text{NO}_3} $$
---
- Reactants: SeCl₆, O₂
- Products: SeO₂, Cl₂
Balancing:
- Start with Se: There is 1 Se atom in SeCl₆, so we need 1 SeO₂ molecule.
- Next, balance Cl: There are 6 Cl atoms in SeCl₆, so we need 3 Cl₂ molecules.
- Finally, balance O: Each SeO₂ has 2 O atoms, so we need 1 O₂ molecule.
Balanced Equation:
$$ \boxed{\text{SeCl}_6 + \text{O}_2 \rightarrow \text{SeO}_2 + 3 \text{Cl}_2} $$
---
$$
\boxed{
\begin{aligned}
1. & \quad 2 \text{Fe} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3 \text{H}_2 \\
2. & \quad 2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 6 \text{H}_2\text{O} + 4 \text{CO}_2 \\
3. & \quad 3 \text{KOH} + \text{H}_3\text{PO}_4 \rightarrow \text{K}_3\text{PO}_4 + 3 \text{H}_2\text{O} \\
4. & \quad \text{SnO}_2 + 2 \text{H}_2 \rightarrow \text{Sn} + 2 \text{H}_2\text{O} \\
5. & \quad 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O} \\
6. & \quad 2 \text{KNO}_3 + \text{H}_2\text{CO}_3 \rightarrow \text{K}_2\text{CO}_3 + 2 \text{HNO}_3 \\
7. & \quad \text{B}_2\text{Br}_6 + 6 \text{HNO}_3 \rightarrow 2 \text{B(NO}_3)_3 + 6 \text{HBr} \\
8. & \quad 2 \text{BF}_3 + 3 \text{Li}_2\text{SO}_3 \rightarrow \text{B}_2(\text{SO}_3)_3 + 6 \text{LiF} \\
9. & \quad 4 (\text{NH}_4)_3\text{PO}_4 + 3 \text{Pb(NO}_3)_4 \rightarrow \text{Pb}_3(\text{PO}_4)_4 + 12 \text{NH}_4\text{NO}_3 \\
10. & \quad \text{SeCl}_6 + \text{O}_2 \rightarrow \text{SeO}_2 + 3 \text{Cl}_2 \\
\end{aligned}
}
$$
---
1. Fe + H₂SO₄ → Fe₂(SO₄)₃ + H₂
- Reactants: Fe, H₂SO₄
- Products: Fe₂(SO₄)₃, H₂
Balancing:
- Start with Fe: There are 2 Fe atoms in Fe₂(SO₄)₃, so we need 2 Fe atoms on the left.
- Next, balance H and S: Each H₂SO₄ provides 2 H atoms and 1 S atom. Since there are 3 SO₄ groups in Fe₂(SO₄)₃, we need 3 H₂SO₄ molecules.
- Finally, balance H₂: From 3 H₂SO₄, we get 6 H atoms, which means we need 3 H₂ molecules.
Balanced Equation:
$$ \boxed{2 \text{Fe} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3 \text{H}_2} $$
---
2. C₂H₆ + O₂ → H₂O + CO₂
- Reactants: C₂H₆, O₂
- Products: H₂O, CO₂
Balancing:
- Start with C: There are 2 C atoms in C₂H₆, so we need 2 CO₂ molecules.
- Next, balance H: There are 6 H atoms in C₂H₆, so we need 3 H₂O molecules.
- Finally, balance O: Each CO₂ has 2 O atoms (total 4 O atoms), and each H₂O has 1 O atom (total 3 O atoms), making a total of 7 O atoms needed. Since O₂ provides 2 O atoms per molecule, we need 7/2 = 3.5 O₂ molecules. To avoid fractions, multiply everything by 2.
Balanced Equation:
$$ \boxed{2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 6 \text{H}_2\text{O} + 4 \text{CO}_2} $$
---
3. KOH + H₃PO₄ → K₃PO₄ + H₂O
- Reactants: KOH, H₃PO₄
- Products: K₃PO₄, H₂O
Balancing:
- Start with K: There are 3 K atoms in K₃PO₄, so we need 3 KOH molecules.
- Next, balance P: There is 1 P atom in H₃PO₄, so we need 1 H₃PO₄ molecule.
- Balance H and O: From 3 KOH, we get 3 H atoms, and from H₃PO₄, we get 3 more H atoms, totaling 6 H atoms, which means we need 3 H₂O molecules.
Balanced Equation:
$$ \boxed{3 \text{KOH} + \text{H}_3\text{PO}_4 \rightarrow \text{K}_3\text{PO}_4 + 3 \text{H}_2\text{O}} $$
---
4. SnO₂ + H₂ → Sn + H₂O
- Reactants: SnO₂, H₂
- Products: Sn, H₂O
Balancing:
- Start with Sn: There is 1 Sn atom in SnO₂, so we need 1 Sn atom on the right.
- Next, balance O: There are 2 O atoms in SnO₂, so we need 2 H₂O molecules.
- Finally, balance H: Each H₂O has 2 H atoms, so we need 2 H₂ molecules.
Balanced Equation:
$$ \boxed{\text{SnO}_2 + 2 \text{H}_2 \rightarrow \text{Sn} + 2 \text{H}_2\text{O}} $$
---
5. NH₃ + O₂ → NO + H₂O
- Reactants: NH₃, O₂
- Products: NO, H₂O
Balancing:
- Start with N: There is 1 N atom in NH₃, so we need 1 NO molecule.
- Next, balance H: There are 3 H atoms in NH₃, so we need 3/2 H₂O molecules. To avoid fractions, multiply everything by 2.
- Balance O: From 2 NO, we have 2 O atoms, and from 3 H₂O, we have 3 O atoms, totaling 5 O atoms. Since O₂ provides 2 O atoms per molecule, we need 5/2 = 2.5 O₂ molecules. Multiply everything by 2 again to eliminate the fraction.
Balanced Equation:
$$ \boxed{4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O}} $$
---
6. KNO₃ + H₂CO₃ → K₂CO₃ + HNO₃
- Reactants: KNO₃, H₂CO₃
- Products: K₂CO₃, HNO₃
Balancing:
- Start with K: There are 2 K atoms in K₂CO₃, so we need 2 KNO₃ molecules.
- Next, balance C: There is 1 C atom in H₂CO₃, so we need 1 H₂CO₃ molecule.
- Balance H and O: From 2 KNO₃, we get 2 HNO₃ molecules, and from H₂CO₃, we get 1 CO₃ group, which matches K₂CO₃.
Balanced Equation:
$$ \boxed{2 \text{KNO}_3 + \text{H}_2\text{CO}_3 \rightarrow \text{K}_2\text{CO}_3 + 2 \text{HNO}_3} $$
---
7. B₂Br₆ + HNO₃ → B(NO₃)₃ + HBr
- Reactants: B₂Br₆, HNO₃
- Products: B(NO₃)₃, HBr
Balancing:
- Start with B: There are 2 B atoms in B₂Br₆, so we need 2 B(NO₃)₃ molecules.
- Next, balance Br: There are 6 Br atoms in B₂Br₆, so we need 6 HBr molecules.
- Balance N and O: Each B(NO₃)₃ has 3 NO₃ groups, so we need 6 HNO₃ molecules.
Balanced Equation:
$$ \boxed{\text{B}_2\text{Br}_6 + 6 \text{HNO}_3 \rightarrow 2 \text{B(NO}_3)_3 + 6 \text{HBr}} $$
---
8. BF₃ + Li₂SO₃ → B₂(SO₃)₃ + LiF
- Reactants: BF₃, Li₂SO₃
- Products: B₂(SO₃)₃, LiF
Balancing:
- Start with B: There are 2 B atoms in B₂(SO₃)₃, so we need 2 BF₃ molecules.
- Next, balance F: Each BF₃ has 3 F atoms, so we need 6 LiF molecules.
- Balance Li and S: Each Li₂SO₃ has 1 S atom, so we need 3 Li₂SO₃ molecules to provide 3 SO₃ groups for B₂(SO₃)₃. This also gives us 6 Li atoms, matching the 6 LiF molecules.
Balanced Equation:
$$ \boxed{2 \text{BF}_3 + 3 \text{Li}_2\text{SO}_3 \rightarrow \text{B}_2(\text{SO}_3)_3 + 6 \text{LiF}} $$
---
9. (NH₄)₃PO₄ + Pb(NO₃)₄ → Pb₃(PO₄)₄ + NH₄NO₃
- Reactants: (NH₄)₃PO₄, Pb(NO₃)₄
- Products: Pb₃(PO₄)₄, NH₄NO₃
Balancing:
- Start with Pb: There are 3 Pb atoms in Pb₃(PO₄)₄, so we need 3 Pb(NO₃)₄ molecules.
- Next, balance P: There are 4 P atoms in Pb₃(PO₄)₄, so we need 4 (NH₄)₃PO₄ molecules.
- Balance N, H, and O: Each (NH₄)₃PO₄ provides 3 NH₄ groups, so we need 12 NH₄NO₃ molecules. Each Pb(NO₃)₄ provides 4 NO₃ groups, so we get 12 NO₃ groups in total, matching the 12 NH₄NO₃ molecules.
Balanced Equation:
$$ \boxed{4 (\text{NH}_4)_3\text{PO}_4 + 3 \text{Pb(NO}_3)_4 \rightarrow \text{Pb}_3(\text{PO}_4)_4 + 12 \text{NH}_4\text{NO}_3} $$
---
10. SeCl₆ + O₂ → SeO₂ + Cl₂
- Reactants: SeCl₆, O₂
- Products: SeO₂, Cl₂
Balancing:
- Start with Se: There is 1 Se atom in SeCl₆, so we need 1 SeO₂ molecule.
- Next, balance Cl: There are 6 Cl atoms in SeCl₆, so we need 3 Cl₂ molecules.
- Finally, balance O: Each SeO₂ has 2 O atoms, so we need 1 O₂ molecule.
Balanced Equation:
$$ \boxed{\text{SeCl}_6 + \text{O}_2 \rightarrow \text{SeO}_2 + 3 \text{Cl}_2} $$
---
Final Answer
$$
\boxed{
\begin{aligned}
1. & \quad 2 \text{Fe} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3 \text{H}_2 \\
2. & \quad 2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 6 \text{H}_2\text{O} + 4 \text{CO}_2 \\
3. & \quad 3 \text{KOH} + \text{H}_3\text{PO}_4 \rightarrow \text{K}_3\text{PO}_4 + 3 \text{H}_2\text{O} \\
4. & \quad \text{SnO}_2 + 2 \text{H}_2 \rightarrow \text{Sn} + 2 \text{H}_2\text{O} \\
5. & \quad 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O} \\
6. & \quad 2 \text{KNO}_3 + \text{H}_2\text{CO}_3 \rightarrow \text{K}_2\text{CO}_3 + 2 \text{HNO}_3 \\
7. & \quad \text{B}_2\text{Br}_6 + 6 \text{HNO}_3 \rightarrow 2 \text{B(NO}_3)_3 + 6 \text{HBr} \\
8. & \quad 2 \text{BF}_3 + 3 \text{Li}_2\text{SO}_3 \rightarrow \text{B}_2(\text{SO}_3)_3 + 6 \text{LiF} \\
9. & \quad 4 (\text{NH}_4)_3\text{PO}_4 + 3 \text{Pb(NO}_3)_4 \rightarrow \text{Pb}_3(\text{PO}_4)_4 + 12 \text{NH}_4\text{NO}_3 \\
10. & \quad \text{SeCl}_6 + \text{O}_2 \rightarrow \text{SeO}_2 + 3 \text{Cl}_2 \\
\end{aligned}
}
$$
Parent Tip: Review the logic above to help your child master the concept of balancing equation practice worksheet.
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/balancing-equation-practice-worksheet/balancing-equation-practice-worksheet-34.jpg)
