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

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 $$\ce{MnF3}$$ ?

• Does this answer your question? Yes find the no. of moles first by dividing each reactant's mass with their molar masses. Then divide with their stoichiometric coefficients to judge the limiting reagent. And do the stoichiometry with that. – Sir Arthur7 Jan 25 '20 at 19:25
• I got 1.21 g of MnF3 as my final answer... is that correct? I also got F2 as my limiting reactant. – Alexa Jan 25 '20 at 19:33
• Could you share your working? If I'm not wrong its something around $\pu{0.45g}$. – Sir Arthur7 Jan 25 '20 at 19:39
• – William R. Ebenezer Jan 25 '20 at 19:40
• No you got it wrong, check out the links by @WilliamR.Ebenezer, the entire process has been worked out clearly. Your calculations are wrong too, for e.g. 1.23/308.74=0.0039 – Sir Arthur7 Jan 25 '20 at 19:50

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

$$F_2$$ 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