Balancing equations worksheet | PDF - Free Printable
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Step-by-step solution for: Balancing equations worksheet | PDF
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Step-by-step solution for: Balancing equations worksheet | PDF
Here are the balanced chemical equations for each problem. I have worked through them step by step to ensure the number of atoms for each element is the same on both sides.
Step-by-Step Balancing:
1. $H_2 + O_2 \rightarrow H_2O$: There are 2 oxygens on the left and 1 on the right. Put a 2 in front of $H_2O$. Now there are 4 hydrogens on the right, so put a 2 in front of $H_2$.
* Result: $2H_2 + 1O_2 \rightarrow 2H_2O$
2. $H_3PO_4 + KOH \rightarrow K_3PO_4 + H_2O$: There are 3 potassiums (K) on the right, so we need 3 KOH on the left. This gives us 3 extra hydrogens from KOH plus the 3 from acid (total 6 H). To get 6 hydrogens on the right, we need 3 water molecules ($3 \times 2 = 6$).
* Result: $1H_3PO_4 + 3KOH \rightarrow 1K_3PO_4 + 3H_2O$
3. $K + B_2O_3 \rightarrow K_2O + B$: Start with Boron (B). There are 2 on the left, so put a 2 in front of B on the right. Now look at Oxygen (O). There are 3 on the left. To get 3 on the right, we technically need 1.5 $K_2O$, but we want whole numbers. Let's try doubling everything later if needed. If we use 3 $K_2O$, we have 3 oxygens. That requires 6 Potassiums (K) on the left.
* Result: $6K + 1B_2O_3 \rightarrow 3K_2O + 2B$
4. $HCl + NaOH \rightarrow NaCl + H_2O$: Count the atoms. 1 H + 1 H = 2 H on left. 2 H on right. 1 Cl left, 1 Cl right. 1 Na left, 1 Na right. 1 O left, 1 O right. It is already balanced.
* Result: $1HCl + 1NaOH \rightarrow 1NaCl + 1H_2O$
5. $Na + NaNO_3 \rightarrow Na_2O + N_2$: This is tricky. Let's balance Nitrogen first. We need 2 $NaNO_3$ to get $N_2$. That gives us 2 Na and 6 O on the left side from the nitrate. On the right, we have $N_2$. We need to balance the 6 oxygens. We need 6 $Na_2O$? No, that makes too much sodium. Let's try balancing Oxygen last.
Let's try: $10Na + 2NaNO_3 \rightarrow 6Na_2O + N_2$.
Left: 12 Na, 2 N, 6 O. Right: 12 Na, 2 N, 6 O. It works.
* Result: $10Na + 2NaNO_3 \rightarrow 6Na_2O + 1N_2$
6. $C + S_8 \rightarrow CS_2$: There are 8 sulfurs on the left. We need 8 sulfurs on the right. Since each $CS_2$ has 2 sulfurs, we need 4 $CS_2$ ($4 \times 2 = 8$). This means we need 4 Carbons on the left.
* Result: $4C + 1S_8 \rightarrow 4CS_2$
7. $Na + O_2 \rightarrow Na_2O_2$: 2 Na on right, so 2 Na on left. 2 O on right, 2 O on left. Already balanced.
* Result: $2Na + 1O_2 \rightarrow 1Na_2O_2$
8. $N_2 + O_2 \rightarrow N_2O_5$: Balance Oxygen first. Left has 2, right has 5. The common multiple is 10. Put a 5 in front of $O_2$ and a 2 in front of $N_2O_5$. Now check Nitrogen. Right has $2 \times 2 = 4$ Nitrogens. Left needs 2 $N_2$.
* Result: $2N_2 + 5O_2 \rightarrow 2N_2O_5$
9. $H_3PO_4 + Mg(OH)_2 \rightarrow Mg_3(PO_4)_2 + H_2O$: Balance Magnesium (Mg) first. Right has 3, so put 3 in front of $Mg(OH)_2$. Balance Phosphate ($PO_4$). Right has 2, so put 2 in front of $H_3PO_4$. Now count Hydrogens. Left: $(2 \times 3) + (3 \times 2) = 6 + 6 = 12$ Hydrogens. Right needs 12 Hydrogens, so put 6 in front of $H_2O$. Check Oxygen (outside phosphate): Left $3 \times 2 = 6$. Right $6 \times 1 = 6$. Balanced.
* Result: $2H_3PO_4 + 3Mg(OH)_2 \rightarrow 1Mg_3(PO_4)_2 + 6H_2O$
10. $NaOH + H_2CO_3 \rightarrow Na_2CO_3 + H_2O$: Balance Sodium (Na). Right has 2, so put 2 in front of NaOH. Now count Hydrogens. Left: 2 (from NaOH) + 2 (from acid) = 4. Right needs 4, so put 2 in front of $H_2O$.
* Result: $2NaOH + 1H_2CO_3 \rightarrow 1Na_2CO_3 + 2H_2O$
11. $KOH + HBr \rightarrow KBr + H_2O$: 1 K, 1 O, 1 H + 1 H, 1 Br. Everything is 1 to 1.
* Result: $1KOH + 1HBr \rightarrow 1KBr + 1H_2O$
12. $H_2 + O_2 \rightarrow H_2O_2$: 2 H left, 2 H right. 2 O left, 2 O right. Already balanced.
* Result: $1H_2 + 1O_2 \rightarrow 1H_2O_2$
13. $Na + O_2 \rightarrow Na_2O$: Balance Oxygen. Left 2, Right 1. Put 2 in front of $Na_2O$. Now Right has 4 Na. Put 4 in front of Na on left.
* Result: $4Na + 1O_2 \rightarrow 2Na_2O$
14. $Al(OH)_3 + H_2CO_3 \rightarrow Al_2(CO_3)_3 + H_2O$: Balance Aluminum (Al). Right has 2, so put 2 in front of $Al(OH)_3$. Balance Carbonate ($CO_3$). Right has 3, so put 3 in front of $H_2CO_3$. Count Hydrogens. Left: $(2 \times 3) + (3 \times 2) = 6 + 6 = 12$. Right needs 12, so put 6 in front of $H_2O$.
* Result: $2Al(OH)_3 + 3H_2CO_3 \rightarrow 1Al_2(CO_3)_3 + 6H_2O$
15. $Al + S_8 \rightarrow Al_2S_3$: Balance Sulfur. Left 8, Right 3. Common multiple is 24. Put 3 in front of $S_8$ ($3 \times 8=24$). Put 8 in front of $Al_2S_3$ ($8 \times 3=24$). Now balance Aluminum. Right has $8 \times 2 = 16$. Put 16 in front of Al.
* Result: $16Al + 3S_8 \rightarrow 8Al_2S_3$
16. $Cs + N_2 \rightarrow Cs_3N$: Balance Nitrogen. Left 2, Right 1. Put 2 in front of $Cs_3N$. Now Right has $2 \times 3 = 6$ Cesiums. Put 6 in front of Cs.
* Result: $6Cs + 1N_2 \rightarrow 2Cs_3N$
17. $Mg + Cl_2 \rightarrow MgCl_2$: 1 Mg left, 1 Mg right. 2 Cl left, 2 Cl right. Already balanced.
* Result: $1Mg + 1Cl_2 \rightarrow 1MgCl_2$
18. $Rb + RbNO_3 \rightarrow Rb_2O + N_2$: Similar to #5. Balance Nitrogen first. Need 2 $RbNO_3$ to get $N_2$. This gives 2 Rb and 6 O on the left (from nitrate). We need 6 O on the right, so put 6 in front of $Rb_2O$. This creates $6 \times 2 = 12$ Rb on the right. We already have 2 Rb from the nitrate on the left, so we need 10 more Rb atoms.
* Result: $10Rb + 2RbNO_3 \rightarrow 6Rb_2O + 1N_2$
19. $C_4H_8 + O_2 \rightarrow CO_2 + H_2O$: Balance Carbon. Left 4, so put 4 in front of $CO_2$. Balance Hydrogen. Left 8, so put 4 in front of $H_2O$. Now count Oxygen on the right: $(4 \times 2) + (4 \times 1) = 8 + 4 = 12$. We need 12 Oxygen on the left. Since $O_2$ has 2, we need 6 $O_2$.
* Result: $1C_4H_8 + 6O_2 \rightarrow 4CO_2 + 4H_2O$
20. $N_2 + H_2 \rightarrow NH_3$: Balance Nitrogen. Left 2, so put 2 in front of $NH_3$. Now Right has $2 \times 3 = 6$ Hydrogens. Left needs 6, so put 3 in front of $H_2$.
* Result: $1N_2 + 3H_2 \rightarrow 2NH_3$
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Final Answer:
1. 2 $H_2$ + 1 $O_2$ $\Rightarrow$ 2 $H_2O$
2. 1 $H_3PO_4$ + 3 $KOH$ $\Rightarrow$ 1 $K_3PO_4$ + 3 $H_2O$
3. 6 $K$ + 1 $B_2O_3$ $\Rightarrow$ 3 $K_2O$ + 2 $B$
4. 1 $HCl$ + 1 $NaOH$ $\Rightarrow$ 1 $NaCl$ + 1 $H_2O$
5. 10 $Na$ + 2 $NaNO_3$ $\Rightarrow$ 6 $Na_2O$ + 1 $N_2$
6. 4 $C$ + 1 $S_8$ $\Rightarrow$ 4 $CS_2$
7. 2 $Na$ + 1 $O_2$ $\Rightarrow$ 1 $Na_2O_2$
8. 2 $N_2$ + 5 $O_2$ $\Rightarrow$ 2 $N_2O_5$
9. 2 $H_3PO_4$ + 3 $Mg(OH)_2$ $\Rightarrow$ 1 $Mg_3(PO_4)_2$ + 6 $H_2O$
10. 2 $NaOH$ + 1 $H_2CO_3$ $\Rightarrow$ 1 $Na_2CO_3$ + 2 $H_2O$
11. 1 $KOH$ + 1 $HBr$ $\Rightarrow$ 1 $KBr$ + 1 $H_2O$
12. 1 $H_2$ + 1 $O_2$ $\Rightarrow$ 1 $H_2O_2$
13. 4 $Na$ + 1 $O_2$ $\Rightarrow$ 2 $Na_2O$
14. 2 $Al(OH)_3$ + 3 $H_2CO_3$ $\Rightarrow$ 1 $Al_2(CO_3)_3$ + 6 $H_2O$
15. 16 $Al$ + 3 $S_8$ $\Rightarrow$ 8 $Al_2S_3$
16. 6 $Cs$ + 1 $N_2$ $\Rightarrow$ 2 $Cs_3N$
17. 1 $Mg$ + 1 $Cl_2$ $\Rightarrow$ 1 $MgCl_2$
18. 10 $Rb$ + 2 $RbNO_3$ $\Rightarrow$ 6 $Rb_2O$ + 1 $N_2$
19. 1 $C_4H_8$ + 6 $O_2$ $\Rightarrow$ 4 $CO_2$ + 4 $H_2O$
20. 1 $N_2$ + 3 $H_2$ $\Rightarrow$ 2 $NH_3$
Step-by-Step Balancing:
1. $H_2 + O_2 \rightarrow H_2O$: There are 2 oxygens on the left and 1 on the right. Put a 2 in front of $H_2O$. Now there are 4 hydrogens on the right, so put a 2 in front of $H_2$.
* Result: $2H_2 + 1O_2 \rightarrow 2H_2O$
2. $H_3PO_4 + KOH \rightarrow K_3PO_4 + H_2O$: There are 3 potassiums (K) on the right, so we need 3 KOH on the left. This gives us 3 extra hydrogens from KOH plus the 3 from acid (total 6 H). To get 6 hydrogens on the right, we need 3 water molecules ($3 \times 2 = 6$).
* Result: $1H_3PO_4 + 3KOH \rightarrow 1K_3PO_4 + 3H_2O$
3. $K + B_2O_3 \rightarrow K_2O + B$: Start with Boron (B). There are 2 on the left, so put a 2 in front of B on the right. Now look at Oxygen (O). There are 3 on the left. To get 3 on the right, we technically need 1.5 $K_2O$, but we want whole numbers. Let's try doubling everything later if needed. If we use 3 $K_2O$, we have 3 oxygens. That requires 6 Potassiums (K) on the left.
* Result: $6K + 1B_2O_3 \rightarrow 3K_2O + 2B$
4. $HCl + NaOH \rightarrow NaCl + H_2O$: Count the atoms. 1 H + 1 H = 2 H on left. 2 H on right. 1 Cl left, 1 Cl right. 1 Na left, 1 Na right. 1 O left, 1 O right. It is already balanced.
* Result: $1HCl + 1NaOH \rightarrow 1NaCl + 1H_2O$
5. $Na + NaNO_3 \rightarrow Na_2O + N_2$: This is tricky. Let's balance Nitrogen first. We need 2 $NaNO_3$ to get $N_2$. That gives us 2 Na and 6 O on the left side from the nitrate. On the right, we have $N_2$. We need to balance the 6 oxygens. We need 6 $Na_2O$? No, that makes too much sodium. Let's try balancing Oxygen last.
Let's try: $10Na + 2NaNO_3 \rightarrow 6Na_2O + N_2$.
Left: 12 Na, 2 N, 6 O. Right: 12 Na, 2 N, 6 O. It works.
* Result: $10Na + 2NaNO_3 \rightarrow 6Na_2O + 1N_2$
6. $C + S_8 \rightarrow CS_2$: There are 8 sulfurs on the left. We need 8 sulfurs on the right. Since each $CS_2$ has 2 sulfurs, we need 4 $CS_2$ ($4 \times 2 = 8$). This means we need 4 Carbons on the left.
* Result: $4C + 1S_8 \rightarrow 4CS_2$
7. $Na + O_2 \rightarrow Na_2O_2$: 2 Na on right, so 2 Na on left. 2 O on right, 2 O on left. Already balanced.
* Result: $2Na + 1O_2 \rightarrow 1Na_2O_2$
8. $N_2 + O_2 \rightarrow N_2O_5$: Balance Oxygen first. Left has 2, right has 5. The common multiple is 10. Put a 5 in front of $O_2$ and a 2 in front of $N_2O_5$. Now check Nitrogen. Right has $2 \times 2 = 4$ Nitrogens. Left needs 2 $N_2$.
* Result: $2N_2 + 5O_2 \rightarrow 2N_2O_5$
9. $H_3PO_4 + Mg(OH)_2 \rightarrow Mg_3(PO_4)_2 + H_2O$: Balance Magnesium (Mg) first. Right has 3, so put 3 in front of $Mg(OH)_2$. Balance Phosphate ($PO_4$). Right has 2, so put 2 in front of $H_3PO_4$. Now count Hydrogens. Left: $(2 \times 3) + (3 \times 2) = 6 + 6 = 12$ Hydrogens. Right needs 12 Hydrogens, so put 6 in front of $H_2O$. Check Oxygen (outside phosphate): Left $3 \times 2 = 6$. Right $6 \times 1 = 6$. Balanced.
* Result: $2H_3PO_4 + 3Mg(OH)_2 \rightarrow 1Mg_3(PO_4)_2 + 6H_2O$
10. $NaOH + H_2CO_3 \rightarrow Na_2CO_3 + H_2O$: Balance Sodium (Na). Right has 2, so put 2 in front of NaOH. Now count Hydrogens. Left: 2 (from NaOH) + 2 (from acid) = 4. Right needs 4, so put 2 in front of $H_2O$.
* Result: $2NaOH + 1H_2CO_3 \rightarrow 1Na_2CO_3 + 2H_2O$
11. $KOH + HBr \rightarrow KBr + H_2O$: 1 K, 1 O, 1 H + 1 H, 1 Br. Everything is 1 to 1.
* Result: $1KOH + 1HBr \rightarrow 1KBr + 1H_2O$
12. $H_2 + O_2 \rightarrow H_2O_2$: 2 H left, 2 H right. 2 O left, 2 O right. Already balanced.
* Result: $1H_2 + 1O_2 \rightarrow 1H_2O_2$
13. $Na + O_2 \rightarrow Na_2O$: Balance Oxygen. Left 2, Right 1. Put 2 in front of $Na_2O$. Now Right has 4 Na. Put 4 in front of Na on left.
* Result: $4Na + 1O_2 \rightarrow 2Na_2O$
14. $Al(OH)_3 + H_2CO_3 \rightarrow Al_2(CO_3)_3 + H_2O$: Balance Aluminum (Al). Right has 2, so put 2 in front of $Al(OH)_3$. Balance Carbonate ($CO_3$). Right has 3, so put 3 in front of $H_2CO_3$. Count Hydrogens. Left: $(2 \times 3) + (3 \times 2) = 6 + 6 = 12$. Right needs 12, so put 6 in front of $H_2O$.
* Result: $2Al(OH)_3 + 3H_2CO_3 \rightarrow 1Al_2(CO_3)_3 + 6H_2O$
15. $Al + S_8 \rightarrow Al_2S_3$: Balance Sulfur. Left 8, Right 3. Common multiple is 24. Put 3 in front of $S_8$ ($3 \times 8=24$). Put 8 in front of $Al_2S_3$ ($8 \times 3=24$). Now balance Aluminum. Right has $8 \times 2 = 16$. Put 16 in front of Al.
* Result: $16Al + 3S_8 \rightarrow 8Al_2S_3$
16. $Cs + N_2 \rightarrow Cs_3N$: Balance Nitrogen. Left 2, Right 1. Put 2 in front of $Cs_3N$. Now Right has $2 \times 3 = 6$ Cesiums. Put 6 in front of Cs.
* Result: $6Cs + 1N_2 \rightarrow 2Cs_3N$
17. $Mg + Cl_2 \rightarrow MgCl_2$: 1 Mg left, 1 Mg right. 2 Cl left, 2 Cl right. Already balanced.
* Result: $1Mg + 1Cl_2 \rightarrow 1MgCl_2$
18. $Rb + RbNO_3 \rightarrow Rb_2O + N_2$: Similar to #5. Balance Nitrogen first. Need 2 $RbNO_3$ to get $N_2$. This gives 2 Rb and 6 O on the left (from nitrate). We need 6 O on the right, so put 6 in front of $Rb_2O$. This creates $6 \times 2 = 12$ Rb on the right. We already have 2 Rb from the nitrate on the left, so we need 10 more Rb atoms.
* Result: $10Rb + 2RbNO_3 \rightarrow 6Rb_2O + 1N_2$
19. $C_4H_8 + O_2 \rightarrow CO_2 + H_2O$: Balance Carbon. Left 4, so put 4 in front of $CO_2$. Balance Hydrogen. Left 8, so put 4 in front of $H_2O$. Now count Oxygen on the right: $(4 \times 2) + (4 \times 1) = 8 + 4 = 12$. We need 12 Oxygen on the left. Since $O_2$ has 2, we need 6 $O_2$.
* Result: $1C_4H_8 + 6O_2 \rightarrow 4CO_2 + 4H_2O$
20. $N_2 + H_2 \rightarrow NH_3$: Balance Nitrogen. Left 2, so put 2 in front of $NH_3$. Now Right has $2 \times 3 = 6$ Hydrogens. Left needs 6, so put 3 in front of $H_2$.
* Result: $1N_2 + 3H_2 \rightarrow 2NH_3$
──────────────────────────────────────
Final Answer:
1. 2 $H_2$ + 1 $O_2$ $\Rightarrow$ 2 $H_2O$
2. 1 $H_3PO_4$ + 3 $KOH$ $\Rightarrow$ 1 $K_3PO_4$ + 3 $H_2O$
3. 6 $K$ + 1 $B_2O_3$ $\Rightarrow$ 3 $K_2O$ + 2 $B$
4. 1 $HCl$ + 1 $NaOH$ $\Rightarrow$ 1 $NaCl$ + 1 $H_2O$
5. 10 $Na$ + 2 $NaNO_3$ $\Rightarrow$ 6 $Na_2O$ + 1 $N_2$
6. 4 $C$ + 1 $S_8$ $\Rightarrow$ 4 $CS_2$
7. 2 $Na$ + 1 $O_2$ $\Rightarrow$ 1 $Na_2O_2$
8. 2 $N_2$ + 5 $O_2$ $\Rightarrow$ 2 $N_2O_5$
9. 2 $H_3PO_4$ + 3 $Mg(OH)_2$ $\Rightarrow$ 1 $Mg_3(PO_4)_2$ + 6 $H_2O$
10. 2 $NaOH$ + 1 $H_2CO_3$ $\Rightarrow$ 1 $Na_2CO_3$ + 2 $H_2O$
11. 1 $KOH$ + 1 $HBr$ $\Rightarrow$ 1 $KBr$ + 1 $H_2O$
12. 1 $H_2$ + 1 $O_2$ $\Rightarrow$ 1 $H_2O_2$
13. 4 $Na$ + 1 $O_2$ $\Rightarrow$ 2 $Na_2O$
14. 2 $Al(OH)_3$ + 3 $H_2CO_3$ $\Rightarrow$ 1 $Al_2(CO_3)_3$ + 6 $H_2O$
15. 16 $Al$ + 3 $S_8$ $\Rightarrow$ 8 $Al_2S_3$
16. 6 $Cs$ + 1 $N_2$ $\Rightarrow$ 2 $Cs_3N$
17. 1 $Mg$ + 1 $Cl_2$ $\Rightarrow$ 1 $MgCl_2$
18. 10 $Rb$ + 2 $RbNO_3$ $\Rightarrow$ 6 $Rb_2O$ + 1 $N_2$
19. 1 $C_4H_8$ + 6 $O_2$ $\Rightarrow$ 4 $CO_2$ + 4 $H_2O$
20. 1 $N_2$ + 3 $H_2$ $\Rightarrow$ 2 $NH_3$
Parent Tip: Review the logic above to help your child master the concept of worksheet on balancing equations.
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