Question

Manganese trifluoride can be prepared by the reaction:
$Mn{l_{2(s)}} + \dfrac{{13}}{2}{F_{2(g)}} \to Mn{F_3} + 2I{F_5}$
What is the minimum amount of ${F_2}$ that must be used to react with $120g$ of $Mn{l_2}$ if only\[75\% \]. ${F_2}$ is utilized to convert all of $Mn{l_2}$ to $Mn{F_3}$.

Answer 1

Hint: As we know that, manganese difluoride reacts with Fluorine and results in the conversion of this compound to manganese trifluoride. We can say that two moles of manganese difluoride would react with thirteen moles of ${F_2}$ to produce two moles of manganese trifluoride and two moles of ${F_5}$.

Complete step-by-step answer:
We know that manganese trifluoride is easily prepared using manganese difluoride and fluorine reacting with each other in the presence of hydrogen fluoride. We are given that:
The yield of $Mn{F_3}$= \[75\% \]
Amount of $Mn{I_2}$= \[120g\]

The reaction given says that two moles of manganese difluoride reacts with thirteen moles of ${F_2}$ to produce two moles of manganese trifluoride and four moles of ${F_5}$ which we can show as follows:
$2Mn{I_2} + 13{F_2} \to 2Mn{F_3} + 4I{F_5}$

Now, we know that the molecular mass of manganese difluoride is \[309g\]and the molecular mass of fluorine is $38g$ and the molecular mass of manganese trifluoride is \[112g\]. So from the above given reaction we can say that:
\[309g\] of $Mn{I_2}$ produces \[112g\] of $Mn{F_3}$.

So, we can say that: $1$ gram of $Mn{I_2}$ will produce $\dfrac{{112}}{{309}}$ grams of $Mn{F_3}$.
Hence, \[120g\] of $Mn{I_2}$ will produce $\dfrac{{112}}{{309}} \times 120 = 43.49$ grams of $Mn{F_3}$.

Now, according to the question, the yield gives \[75\% \].
So, the actual mass of $Mn{F_3}$ obtained on the reaction is given as $\dfrac{{75}}{{100}} \times 43.49 = 32.62$ grams of $Mn{F_3}$.

Now, we already know that \[112g\]of $Mn{F_3}$ is produced from \[38g\] of ${F_2}$.
Therefore, we can calculate the amount of fluorine produced from $32.62g$ using the unitary method as:
$1$ gram of $Mn{F_3}$ is produced from $\dfrac{{38}}{{112}}$ grams of ${F_2}$.
Hence, \[32.62g\]of $Mn{F_3}$ is produced from $\dfrac{{38}}{{112}} \times 32.62 = 12.78$ grams of ${F_2}$. Thus, the minimum amount of ${F_2}$ that must be used to react with $Mn{l_2}$ if only $75\% $. ${F_2}$ is utilized to convert all of $Mn{l_2}$ to $Mn{F_3}$ is \[12.78g\].

Hence, the correct answer is \[12.78g\].

Note: Remember that Manganese trifluoride is a fluorinating agent and can be generally prepared by treating manganese difluoride with fluorine in the presence of hydrogen fluoride. Manganese trifluoride is basically a Lewis acid and results in the formation of a variety of derivatives. This inorganic compound is very useful in converting hydrocarbons into the fluorocarbons. Manganic fluoride is the another name of manganese trifluoride.