Worksheet for practicing balancing chemical equations and writing chemical reactions from word descriptions.
Balancing Chemical Equations worksheet with problems and instructions for students to balance chemical equations and write balanced equations from word descriptions.
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Step-by-step solution for: Free Printable Balancing Equations Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Free Printable Balancing Equations Worksheets
Here are the balanced chemical equations for your homework.
1) $\text{CaC}_2 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + \text{Ca(OH)}_2$
* Step 1: Count atoms on both sides.
* Left: Ca=1, C=2, H=2, O=1
* Right: Ca=1, C=2, H=4 (2+2), O=2
* Step 2: Balance Oxygen and Hydrogen. We need 2 water molecules on the left to match the right side.
* Try: $\text{CaC}_2 + \mathbf{2}\text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + \text{Ca(OH)}_2$
* Check:
* Left: Ca=1, C=2, H=4, O=2
* Right: Ca=1, C=2, H=4, O=2
* Answer: $1, 2, 1, 1$
2) $\text{Ca(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{O}$
* Step 1: Balance Calcium (Ca). There are 3 on the right, so put a 3 on the left.
* $3\text{Ca(OH)}_2 + \dots$
* Step 2: Balance Phosphate ($\text{PO}_4$). There are 2 groups on the right, so put a 2 on the left.
* $\dots + 2\text{H}_3\text{PO}_4 \rightarrow \dots$
* Step 3: Count Hydrogens.
* Left: $(3 \times 2) + (2 \times 3) = 6 + 6 = 12$ Hydrogens.
* Right: We need 12 Hydrogens in water ($\text{H}_2\text{O}$). Since each water has 2 H, we need 6 waters.
* Answer: $3, 2, 1, 6$
3) $\text{SrCO}_3 + \text{NH}_4\text{Br} \rightarrow \text{SrBr}_2 + (\text{NH}_4)_2\text{CO}_3$
* Step 1: Balance Bromine (Br). There are 2 on the right, so put a 2 on the left.
* $\dots + 2\text{NH}_4\text{Br} \rightarrow \dots$
* Step 2: Check Ammonium ($\text{NH}_4$). Now there are 2 on the left, which matches the 2 on the right.
* Check: Sr=1, C=1, O=3, N=2, H=8, Br=2 on both sides.
* Answer: $1, 2, 1, 1$
4) $\text{C}_7\text{H}_{16} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$
* Step 1: Balance Carbon (C). 7 on left $\rightarrow$ 7 on right.
* $\dots \rightarrow 7\text{CO}_2 + \dots$
* Step 2: Balance Hydrogen (H). 16 on left $\rightarrow$ need 8 waters on right ($8 \times 2 = 16$).
* $\dots \rightarrow \dots + 8\text{H}_2\text{O}$
* Step 3: Balance Oxygen (O).
* Right side total: $(7 \times 2) + (8 \times 1) = 14 + 8 = 22$ Oxygens.
* Left side needs 22. Since $\text{O}_2$ has 2 atoms, we need 11 molecules.
* Answer: $1, 11, 7, 8$
5) $\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$
* Step 1: Balance Carbon. 4 on left $\rightarrow$ 4 on right.
* Step 2: Balance Hydrogen. 10 on left $\rightarrow$ 5 waters on right.
* Step 3: Balance Oxygen. Right side: $(4 \times 2) + 5 = 13$.
* We need 13 oxygens on the left. That would be $6.5 \text{O}_2$.
* To remove the decimal, multiply everything by 2.
* Answer: $2, 13, 8, 10$
6) $\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3$
* Step 1: Balance Iron (Fe). 2 on right $\rightarrow$ 2 on left.
* Step 2: Balance Oxygen. 3 on right, 2 on left. The common multiple is 6.
* Put 3 in front of $\text{O}_2$ (makes 6).
* Put 2 in front of $\text{Fe}_2\text{O}_3$ (makes 6).
* Step 3: Re-balance Iron. Now we have 4 Fe on the right ($2 \times 2$), so we need 4 Fe on the left.
* Answer: $4, 3, 2$
---
a) Hydrogen plus oxygen yields water.
* Unbalanced: $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$
* Balance O: Need 2 waters. $\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$
* Balance H: Need 2 hydrogens on left.
* Answer: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$
b) Sodium plus water yields sodium hydroxide plus hydrogen.
* Unbalanced: $\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2$
* Balance H: Left has 2, Right has 3 ($1+2$). Let's double the water to get even numbers.
* $\text{Na} + 2\text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2$
* Now Right has 4 H total? No, let's look at OH.
* Let's try putting a 2 in front of NaOH.
* $\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$
* Check H: Left=4, Right=2+2=4. Good.
* Check Na: Right=2, so Left needs 2.
* Answer: $2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$
c) Hydrochloric acid plus iron (III) sulfide yields iron (III) chloride plus hydrogen sulfide.
* Formulas: $\text{HCl} + \text{Fe}_2\text{S}_3 \rightarrow \text{FeCl}_3 + \text{H}_2\text{S}$
* Balance Fe: 2 on left $\rightarrow$ 2 on right ($2\text{FeCl}_3$).
* Balance S: 3 on left $\rightarrow$ 3 on right ($3\text{H}_2\text{S}$).
* Balance Cl: Right has $2 \times 3 = 6$. Left needs 6 ($6\text{HCl}$).
* Check H: Left 6, Right $3 \times 2 = 6$. Good.
* Answer: $6\text{HCl} + \text{Fe}_2\text{S}_3 \rightarrow 2\text{FeCl}_3 + 3\text{H}_2\text{S}$
d) Aluminum sulfate plus calcium hydroxide yields aluminum hydroxide plus calcium sulfate.
* Formulas: $\text{Al}_2(\text{SO}_4)_3 + \text{Ca(OH)}_2 \rightarrow \text{Al(OH)}_3 + \text{CaSO}_4$
* Balance Al: 2 on left $\rightarrow$ 2 on right ($2\text{Al(OH)}_3$).
* Balance Sulfate ($\text{SO}_4$): 3 on left $\rightarrow$ 3 on right ($3\text{CaSO}_4$).
* Balance Ca: Right has 3, so Left needs 3 ($3\text{Ca(OH)}_2$).
* Check OH: Left $3 \times 2 = 6$. Right $2 \times 3 = 6$. Good.
* Answer: $\text{Al}_2(\text{SO}_4)_3 + 3\text{Ca(OH)}_2 \rightarrow 2\text{Al(OH)}_3 + 3\text{CaSO}_4$
e) Zinc and lead (II) nitrate react to form zinc nitrate and lead.
* Formulas: $\text{Zn} + \text{Pb(NO}_3)_2 \rightarrow \text{Zn(NO}_3)_2 + \text{Pb}$
* Note: Zinc usually forms a +2 ion ($\text{Zn}^{2+}$), same as Lead (II). The nitrates swap places directly.
* Check atoms: Zn=1, Pb=1, N=2, O=6 on both sides. It is already balanced.
* Answer: $\text{Zn} + \text{Pb(NO}_3)_2 \rightarrow \text{Zn(NO}_3)_2 + \text{Pb}$
f) Aluminum bromide and chlorine gas react to form aluminum chloride and bromine gas.
* Formulas: $\text{AlBr}_3 + \text{Cl}_2 \rightarrow \text{AlCl}_3 + \text{Br}_2$
* Balance Br: Left 3, Right 2. Common multiple is 6.
* Put 2 in front of $\text{AlBr}_3$.
* Put 3 in front of $\text{Br}_2$.
* $2\text{AlBr}_3 + \text{Cl}_2 \rightarrow \text{AlCl}_3 + 3\text{Br}_2$
* Balance Al: Left 2, so Right needs 2 ($2\text{AlCl}_3$).
* Balance Cl: Right $2 \times 3 = 6$. Left needs 6, so put 3 in front of $\text{Cl}_2$.
* Answer: $2\text{AlBr}_3 + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Br}_2$
Final Answer:
Balance the following:
1) $1, 2, 1, 1$
2) $3, 2, 1, 6$
3) $1, 2, 1, 1$
4) $1, 11, 7, 8$
5) $2, 13, 8, 10$
6) $4, 3, 2$
Write and balance:
a) $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$
b) $2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$
c) $6\text{HCl} + \text{Fe}_2\text{S}_3 \rightarrow 2\text{FeCl}_3 + 3\text{H}_2\text{S}$
d) $\text{Al}_2(\text{SO}_4)_3 + 3\text{Ca(OH)}_2 \rightarrow 2\text{Al(OH)}_3 + 3\text{CaSO}_4$
e) $\text{Zn} + \text{Pb(NO}_3)_2 \rightarrow \text{Zn(NO}_3)_2 + \text{Pb}$
f) $2\text{AlBr}_3 + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Br}_2$
Part 1: Balance the following chemical equations
1) $\text{CaC}_2 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + \text{Ca(OH)}_2$
* Step 1: Count atoms on both sides.
* Left: Ca=1, C=2, H=2, O=1
* Right: Ca=1, C=2, H=4 (2+2), O=2
* Step 2: Balance Oxygen and Hydrogen. We need 2 water molecules on the left to match the right side.
* Try: $\text{CaC}_2 + \mathbf{2}\text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + \text{Ca(OH)}_2$
* Check:
* Left: Ca=1, C=2, H=4, O=2
* Right: Ca=1, C=2, H=4, O=2
* Answer: $1, 2, 1, 1$
2) $\text{Ca(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{O}$
* Step 1: Balance Calcium (Ca). There are 3 on the right, so put a 3 on the left.
* $3\text{Ca(OH)}_2 + \dots$
* Step 2: Balance Phosphate ($\text{PO}_4$). There are 2 groups on the right, so put a 2 on the left.
* $\dots + 2\text{H}_3\text{PO}_4 \rightarrow \dots$
* Step 3: Count Hydrogens.
* Left: $(3 \times 2) + (2 \times 3) = 6 + 6 = 12$ Hydrogens.
* Right: We need 12 Hydrogens in water ($\text{H}_2\text{O}$). Since each water has 2 H, we need 6 waters.
* Answer: $3, 2, 1, 6$
3) $\text{SrCO}_3 + \text{NH}_4\text{Br} \rightarrow \text{SrBr}_2 + (\text{NH}_4)_2\text{CO}_3$
* Step 1: Balance Bromine (Br). There are 2 on the right, so put a 2 on the left.
* $\dots + 2\text{NH}_4\text{Br} \rightarrow \dots$
* Step 2: Check Ammonium ($\text{NH}_4$). Now there are 2 on the left, which matches the 2 on the right.
* Check: Sr=1, C=1, O=3, N=2, H=8, Br=2 on both sides.
* Answer: $1, 2, 1, 1$
4) $\text{C}_7\text{H}_{16} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$
* Step 1: Balance Carbon (C). 7 on left $\rightarrow$ 7 on right.
* $\dots \rightarrow 7\text{CO}_2 + \dots$
* Step 2: Balance Hydrogen (H). 16 on left $\rightarrow$ need 8 waters on right ($8 \times 2 = 16$).
* $\dots \rightarrow \dots + 8\text{H}_2\text{O}$
* Step 3: Balance Oxygen (O).
* Right side total: $(7 \times 2) + (8 \times 1) = 14 + 8 = 22$ Oxygens.
* Left side needs 22. Since $\text{O}_2$ has 2 atoms, we need 11 molecules.
* Answer: $1, 11, 7, 8$
5) $\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$
* Step 1: Balance Carbon. 4 on left $\rightarrow$ 4 on right.
* Step 2: Balance Hydrogen. 10 on left $\rightarrow$ 5 waters on right.
* Step 3: Balance Oxygen. Right side: $(4 \times 2) + 5 = 13$.
* We need 13 oxygens on the left. That would be $6.5 \text{O}_2$.
* To remove the decimal, multiply everything by 2.
* Answer: $2, 13, 8, 10$
6) $\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3$
* Step 1: Balance Iron (Fe). 2 on right $\rightarrow$ 2 on left.
* Step 2: Balance Oxygen. 3 on right, 2 on left. The common multiple is 6.
* Put 3 in front of $\text{O}_2$ (makes 6).
* Put 2 in front of $\text{Fe}_2\text{O}_3$ (makes 6).
* Step 3: Re-balance Iron. Now we have 4 Fe on the right ($2 \times 2$), so we need 4 Fe on the left.
* Answer: $4, 3, 2$
---
Part 2: Write and balance the following chemical equations
a) Hydrogen plus oxygen yields water.
* Unbalanced: $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$
* Balance O: Need 2 waters. $\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$
* Balance H: Need 2 hydrogens on left.
* Answer: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$
b) Sodium plus water yields sodium hydroxide plus hydrogen.
* Unbalanced: $\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2$
* Balance H: Left has 2, Right has 3 ($1+2$). Let's double the water to get even numbers.
* $\text{Na} + 2\text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2$
* Now Right has 4 H total? No, let's look at OH.
* Let's try putting a 2 in front of NaOH.
* $\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$
* Check H: Left=4, Right=2+2=4. Good.
* Check Na: Right=2, so Left needs 2.
* Answer: $2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$
c) Hydrochloric acid plus iron (III) sulfide yields iron (III) chloride plus hydrogen sulfide.
* Formulas: $\text{HCl} + \text{Fe}_2\text{S}_3 \rightarrow \text{FeCl}_3 + \text{H}_2\text{S}$
* Balance Fe: 2 on left $\rightarrow$ 2 on right ($2\text{FeCl}_3$).
* Balance S: 3 on left $\rightarrow$ 3 on right ($3\text{H}_2\text{S}$).
* Balance Cl: Right has $2 \times 3 = 6$. Left needs 6 ($6\text{HCl}$).
* Check H: Left 6, Right $3 \times 2 = 6$. Good.
* Answer: $6\text{HCl} + \text{Fe}_2\text{S}_3 \rightarrow 2\text{FeCl}_3 + 3\text{H}_2\text{S}$
d) Aluminum sulfate plus calcium hydroxide yields aluminum hydroxide plus calcium sulfate.
* Formulas: $\text{Al}_2(\text{SO}_4)_3 + \text{Ca(OH)}_2 \rightarrow \text{Al(OH)}_3 + \text{CaSO}_4$
* Balance Al: 2 on left $\rightarrow$ 2 on right ($2\text{Al(OH)}_3$).
* Balance Sulfate ($\text{SO}_4$): 3 on left $\rightarrow$ 3 on right ($3\text{CaSO}_4$).
* Balance Ca: Right has 3, so Left needs 3 ($3\text{Ca(OH)}_2$).
* Check OH: Left $3 \times 2 = 6$. Right $2 \times 3 = 6$. Good.
* Answer: $\text{Al}_2(\text{SO}_4)_3 + 3\text{Ca(OH)}_2 \rightarrow 2\text{Al(OH)}_3 + 3\text{CaSO}_4$
e) Zinc and lead (II) nitrate react to form zinc nitrate and lead.
* Formulas: $\text{Zn} + \text{Pb(NO}_3)_2 \rightarrow \text{Zn(NO}_3)_2 + \text{Pb}$
* Note: Zinc usually forms a +2 ion ($\text{Zn}^{2+}$), same as Lead (II). The nitrates swap places directly.
* Check atoms: Zn=1, Pb=1, N=2, O=6 on both sides. It is already balanced.
* Answer: $\text{Zn} + \text{Pb(NO}_3)_2 \rightarrow \text{Zn(NO}_3)_2 + \text{Pb}$
f) Aluminum bromide and chlorine gas react to form aluminum chloride and bromine gas.
* Formulas: $\text{AlBr}_3 + \text{Cl}_2 \rightarrow \text{AlCl}_3 + \text{Br}_2$
* Balance Br: Left 3, Right 2. Common multiple is 6.
* Put 2 in front of $\text{AlBr}_3$.
* Put 3 in front of $\text{Br}_2$.
* $2\text{AlBr}_3 + \text{Cl}_2 \rightarrow \text{AlCl}_3 + 3\text{Br}_2$
* Balance Al: Left 2, so Right needs 2 ($2\text{AlCl}_3$).
* Balance Cl: Right $2 \times 3 = 6$. Left needs 6, so put 3 in front of $\text{Cl}_2$.
* Answer: $2\text{AlBr}_3 + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Br}_2$
Final Answer:
Balance the following:
1) $1, 2, 1, 1$
2) $3, 2, 1, 6$
3) $1, 2, 1, 1$
4) $1, 11, 7, 8$
5) $2, 13, 8, 10$
6) $4, 3, 2$
Write and balance:
a) $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$
b) $2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$
c) $6\text{HCl} + \text{Fe}_2\text{S}_3 \rightarrow 2\text{FeCl}_3 + 3\text{H}_2\text{S}$
d) $\text{Al}_2(\text{SO}_4)_3 + 3\text{Ca(OH)}_2 \rightarrow 2\text{Al(OH)}_3 + 3\text{CaSO}_4$
e) $\text{Zn} + \text{Pb(NO}_3)_2 \rightarrow \text{Zn(NO}_3)_2 + \text{Pb}$
f) $2\text{AlBr}_3 + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Br}_2$
Parent Tip: Review the logic above to help your child master the concept of chemical equation worksheet.
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/chemical-equation-worksheet/chemical-equation-worksheet-4.jpg)
