Finding mass of MnF3 is produced in a reaction
Here we have an already balanced equation given in the question-

$$\ce{2 MnI2 + 13 F2⟶ 2 MnF3 + 4 IF5}$$

We find the given number of moles of each reactant by dividing their given masses by their respective molar masses

Given mass of $\ce{MnI2}=\pu{1.23g}$
Molar mass of $\ce{MnI2}= \pu{308.74 g/mol}$

Amount of $\ce{MnI2} = \frac{1.23}{308.74}= \pu{0.0039mol}$

Given mass of $\ce{F2}=\pu{25g}$
Molar mass of $\ce{F2}= \pu{38 g/mol}$

Amount of $\ce{F2} = \frac{25}{38}= \pu{0.657mol}$

Now we find the mole ratio from the given information and compare the calculated ratio to the actual ratio. 
If more than $\pu{13 moles}$ of $\ce{F2}$ are available per $\pu{2 moles}$ of $\ce{MnI2}$, the $\ce{F2}$ is in excess and $\ce{MnI2}$ is the limiting reactant. If less than $\pu{13 moles}$ of $\ce{F2}$ are available per $\pu{2 moles}$ of $\ce{MnI2}$, $\ce{F2}$ is the limiting reactant.

Let's say that all of the $\pu{0.657 moles}$ of $\ce{F2}$ were to be used up, there would need to be 

$\frac{2}{13}\times0.657=\pu{0.101  moles} \text{ of } \ce{MnI2}$. 

But there is only $\pu{0.0039 moles} \text{ of }\ce{ MnI2}$ available which makes it the limiting reactant.

If all of the $\pu{0.0039 moles} \text{ of }\ce{ MnI2}$ were used up, there would need to be

$0.0039\times \frac{13}{2} = \pu{0.0254 moles} \text{ of } \ce{F2}$.

Because there is an excess of $\ce{F2}$, the $\ce{ MnI2}$ amount is used to calculate the amount of the products in the reaction i.e. amount of $\ce{MnF3}$ produced.

From the equation given its easy to observe that the stoichiometric ratio of $\ce{MnI2}$ and $\ce{MnF3} = 1:1$

Thus amount of $\ce{MnF3}$ produced = $\pu{0.0039 mol}$
Molar mass of $\ce{MnF3} = \pu{111.94 g/mol}$

Mass of $\ce{MnF3} = 111.94 \times 0.00398 = \pu{0.445g}$