mgf₂ + li₂o₃ → mgo₃ + lif
**hint** you only need to add coefficients where the orange boxes are!

Question

Accepted Answer

# Explanation:
## Step1: Count atoms before balancing
- For $Mg$: In $MgF_2$, there is $1$ $Mg$ atom. In $Li_2O_3$, there is $0$ $Mg$ atoms. So before balancing, $Mg = 1$.
- For $F$: In $MgF_2$, there are $2$ $F$ atoms. In $Li_2O_3$, there are $0$ $F$ atoms. So before balancing, $F=2$.
- For $Li$: In $MgF_2$, there are $0$ $Li$ atoms. In $Li_2O_3$, there are $2$ $Li$ atoms. So before balancing, $Li = 2$.
- For $O$: In $MgF_2$, there are $0$ $O$ atoms. In $Li_2O_3$, there are $3$ $O$ atoms. So before balancing, $O=3$.

## Step2: Balance the equation
The balanced chemical equation is $MgF_2+Li_2O_3 = MgO_3 + 2LiF$
- After balancing for $Mg$: In $MgO_3$, there is $1$ $Mg$ atom.
- After balancing for $F$: In $2LiF$, there are $2$ $F$ atoms.
- After balancing for $Li$: In $2LiF$, there are $2$ $Li$ atoms.
- After balancing for $O$: In $MgO_3$, there are $3$ $O$ atoms.

# Answer:
The coefficient in the box is $2$.
| Element | # of atoms before balancing | # of atoms AFTER balancing |
| ---- | ---- | ---- |
| $Mg$ | $1$ | $1$ |
| $F$ | $2$ | $2$ |
| $Li$ | $2$ | $2$ |
| $O$ | $3$ | $3$ |

Element	# of atoms before balancing	# of atoms AFTER balancing
\(F\)	\(2\)	\(2\)
\(Li\)	\(2\)	\(2\)
\(O\)	\(3\)	\(3\)

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Question

Explanation:

Step1: Count atoms before balancing

Step2: Balance the equation

Answer: