'Manganese(III) fluoride, MnF3, can be prepared by the following reaction: 2MnI2(s) + 13F2(g) ? 2MnF3(s) + 4IF5(l) If the percentage yield of MnF3 is always approximately 56%, how many grams of MnF3 can be expected if 10.0 grams of each reactant is used in an experiment?'

Answer 1

This is a limiting reagent (LR) problem and you know that because an amount is given for each reactant (in this cse 10 g each).
2MnI2(s) + 13F2(g)--> 2MnF3(s) + 4IF5(l)

mols MnI2 = grams/molar mass = ?
mols F2 = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols MnI2 to mols MnF3.
Do the same to convert mols F2 to mols MnF3.

It is likely that the two values will not agree which means one is not right. The correct value in LR problems is ALWAYS the smaller value and the reagent responsible for producing that amount is the LR.

Using the smaller amount convert mols to grams. grams = mols x molar mass.
That is the 100% (theoretical yield). To find the yield ad 56%. then
theoretical yield x 0.56 = ?

Answer 2

mols MnI2=0.03246
mols F2=0.26315

Answer 3

.0324*2/2=0.0346 mols MnI2
0.26315*2/13=0.020242 mols F2
F2 =LR
.0202 mols MnF3x112g/mol MnF3=2.26g MnF3

Answer 4

3.875g MnF3

Answer 5

I don't buy all of this.
.0324*2/2=0.0346 mols MnI2
I don't know how you did this but 0.0324 x 2/2 should be 0.0324.

0.26315*2/13=0.020242 mols F2
0.2631 x 2/13 = 0.0404 so MnI2 is the LR. right?

F2 =LR
.0202 mols MnF3x112g/mol MnF3=2.26g MnF3