Finding mass of MnF3 is produced in a reaction [closed]








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The question is as follows:



> 
> If $\pu{1.23 g}$ of $\ce{MnI2}$ reacts with $\pu{25.0 g}$ of $\ce{F2}$, what mass of $\ce{MnF3}$ is produced?
> 
> 
> 


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


I think this is a limiting reactant problem but I'm having difficulties with starting the question off. Do I find the mass of both $\ce{MnI2}$ $\pu{(308.74 g/mol)}$ and $\ce{F2}$ $\pu{(38g/mol)}$ and divide them both by their given masses $\pu{1.23 g}$ and $\pu{25.0 g}$ respectively?


Once I do that, do I find the limiting and excess reactant, take the limiting reactant and do stoichiometry to find the mass of $\ce{MnF3}$ ?




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


 2

$F\_2$ is present in great excess. Do the calculation with the manganese compounds.


$1.23$ g $MnI\_2$ contains $1.23 g /308.74$ = $0.004$ mol $MnI\_2$. The reaction produces the same number of moles of $MnF\_3$. As the molar mass of $MnF\_3$ is $55 + 3·19 = 112$ g/mol, $0.004$ mol $MnF\_3$ weighs $0.004·112$ g = $0.448$ g


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