Free Printable Balancing Equations Worksheets - Free Printable

Here are the balanced chemical equations for your worksheet. I have worked through each one to make sure the number of atoms for every element is the same on both sides.

i. $2\text{NaF} \rightarrow 2\text{Na} + \text{F}_2$
*(Reasoning: You need 2 Fluorines on the left to match the $\text{F}_2$ on the right. This gives you 2 Sodiums, so you put a 2 in front of Na.)*

ii. $6\text{Li} + \text{N}_2 \rightarrow 2\text{Li}_3\text{N}$
*(Reasoning: There are 2 Nitrogens on the left, so you need 2 $\text{Li}_3\text{N}$ on the right. That makes 6 Lithiums total ($2 \times 3$), so you need 6 Li on the left.)*

iii. $2\text{LiBr} + \text{F}_2 \rightarrow 2\text{LiF} + \text{Br}_2$
*(Reasoning: You need 2 Bromines on the left to match $\text{Br}_2$. This gives 2 Lithiums, so you need 2 LiF on the right. Now Fluorine is also balanced with 2 on each side.)*

iv. $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$
*(Reasoning: Start with 2 Nitrogens on the left, so make 2 $\text{NH}_3$ on the right. That creates 6 Hydrogens ($2 \times 3$). To get 6 Hydrogens on the left, you need 3 $\text{H}_2$.)*

v. $\text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}$
*(Reasoning: Balance Carbon first (2 on left $\rightarrow$ 2 $\text{CO}_2$). Balance Hydrogen next (4 on left $\rightarrow$ 2 $\text{H}_2\text{O}$). Finally, count Oxygen: $4 + 2 = 6$ on the right, so you need 3 $\text{O}_2$ on the left.)*

vi. $\text{Al}_2\text{S}_3 \rightarrow 2\text{Al} + 3\text{S}$
*(Reasoning: The subscript 2 on Al becomes a coefficient of 2. The subscript 3 on S becomes a coefficient of 3.)*

vii. $4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$
*(Reasoning: This is tricky because of the odd/even numbers. If you put a 2 in front of $\text{Al}_2\text{O}_3$, you get 4 Al and 6 O. To get 6 O on the left, you need 3 $\text{O}_2$. To get 4 Al on the left, you need 4 Al.)*

viii. $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$
*(Reasoning: If you just put a 1 in front of water, you have an odd number of oxygens. By putting a 2 in front of $\text{H}_2\text{O}_2$ and $\text{H}_2\text{O}$, you get 4 H and 4 O on the left, and 4 H and 4 O ($2+2$) on the right.)*

ix. $\text{NH}_4\text{NO}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}$
*(Reasoning: There are 2 Nitrogens on the left, which matches $\text{N}_2$. There are 4 Hydrogens on the left, so you need 2 $\text{H}_2\text{O}$ to get 4 Hydrogens on the right. Oxygen is now balanced with 2 on each side.)*

x. $2\text{K} + \text{Cl}_2 \rightarrow 2\text{KCl}$
*(Reasoning: You have 2 Chlorines on the left, so you need 2 KCl on the right. This gives 2 Potassiums, so you need 2 K on the left.)*

xi. $2\text{NaCl} \rightarrow 2\text{Na} + \text{Cl}_2$
*(Reasoning: You need 2 Chlorines on the left to match $\text{Cl}_2$. This gives 2 Sodiums, so you put a 2 in front of Na.)*

xii. $3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2$
*(Reasoning: Balance Iron first (3 on right $\rightarrow$ 3 Fe). Balance Oxygen next (4 on right $\rightarrow$ 4 $\text{H}_2\text{O}$). Finally, balance Hydrogen: $4 \times 2 = 8$ on the left, so you need 4 $\text{H}_2$ on the right.)*

xiii. $2\text{N}_2 + 5\text{O}_2 \rightarrow 2\text{N}_2\text{O}_5$
*(Reasoning: Similar to problem vii. Put a 2 in front of the product to get even numbers: 4 N and 10 O. To get 4 N on the left, use 2 $\text{N}_2$. To get 10 O on the left, use 5 $\text{O}_2$.)*

xiv. $2\text{KBr} + \text{Cl}_2 \rightarrow 2\text{KCl} + \text{Br}_2$
*(Reasoning: You need 2 Bromines on the left to match $\text{Br}_2$. This gives 2 Potassiums, so you need 2 KCl on the right. Chlorine is now balanced with 2 on each side.)*

xv. $4\text{P} + 3\text{O}_2 \rightarrow \text{P}_4\text{O}_6$
*(Reasoning: There are 4 Phosphorus atoms on the right, so you need 4 P on the left. There are 6 Oxygen atoms on the right, so you need 3 $\text{O}_2$ on the left ($3 \times 2 = 6$).)*

Final Answer:
i. 2NaF $\rightarrow$ 2Na + F$_2$
ii. 6Li + N$_2$ $\rightarrow$ 2Li$_3$N
iii. 2LiBr + F$_2$ $\rightarrow$ 2LiF + Br$_2$
iv. N$_2$ + 3H$_2$ $\rightarrow$ 2NH$_3$
v. C$_2$H$_4$ + 3O$_2$ $\rightarrow$ 2CO$_2$ + 2H$_2$O
vi. Al$_2$S$_3$ $\rightarrow$ 2Al + 3S
vii. 4Al + 3O$_2$ $\rightarrow$ 2Al$_2$O$_3$
viii. 2H$_2$O$_2$ $\rightarrow$ 2H$_2$O + O$_2$
ix. NH$_4$NO$_2$ $\rightarrow$ N$_2$ + 2H$_2$O
x. 2K + Cl$_2$ $\rightarrow$ 2KCl
xi. 2NaCl $\rightarrow$ 2Na + Cl$_2$
xii. 3Fe + 4H$_2$O $\rightarrow$ Fe$_3$O$_4$ + 4H$_2$
xiii. 2N$_2$ + 5O$_2$ $\rightarrow$ 2N$_2$O$_5$
xiv. 2KBr + Cl$_2$ $\rightarrow$ 2KCl + Br$_2$
xv. 4P + 3O$_2$ $\rightarrow$ P$_4$O$_6$