'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?'
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 = ?
mols MnI2=0.03246
mols F2=0.26315
3.875g MnF3
To find out how many grams of MnF3 can be expected if 10.0 grams of each reactant is used in the experiment, we need to calculate the theoretical yield first.
The balanced equation for the reaction is:
2MnI2(s) + 13F2(g) → 2MnF3(s) + 4IF5(l)
From the equation, we can see that the ratio of MnI2 to MnF3 is 2:2, and the ratio of F2 to MnF3 is 13:2.
Step 1: Calculate the moles of MnI2:
Molar mass of MnI2 = atomic mass of Mn (55.845 g/mol) + 2 * atomic mass of I (2 * 126.9045 g/mol)
Molar mass of MnI2 = 55.845 g/mol + 253.809 g/mol = 309.654 g/mol
Moles of MnI2 = mass of MnI2 / molar mass of MnI2
Moles of MnI2 = 10.0 g / 309.654 g/mol
Step 2: Calculate the moles of F2:
Moles of F2 = mass of F2 / molar mass of F2
Moles of F2 = 10.0 g / molar mass of F2
Step 3: Compare the moles of MnI2 and F2 to find the limiting reactant:
The limiting reactant is the one that is completely consumed in the reaction and determines the amount of product formed.
Based on the balanced equation, the ratio of MnI2 to F2 in moles is 2:13.
Divide the moles of MnI2 by 2 and divide the moles of F2 by 13 to see which is smaller.
Moles of MnF3 = (Moles of MnI2) / 2
Moles of MnF3 = (10.0 g / 309.654 g/mol) / 2
Moles of MnF3 = (Moles of F2) / 13
Moles of MnF3 = (10.0 g / molar mass of F2) / 13
If the moles of MnF3 calculated from MnI2 are smaller than the moles of MnF3 calculated from F2, then MnI2 is the limiting reactant. Otherwise, F2 is the limiting reactant.
Step 4: Calculate the theoretical yield of MnF3:
The theoretical yield is the maximum amount of product that can be obtained from the limiting reactant.
If MnI2 is the limiting reactant:
Moles of MnF3 = (10.0 g / 309.654 g/mol) / 2
If F2 is the limiting reactant:
Moles of MnF3 = (10.0 g / molar mass of F2) / 13
Step 5: Convert the moles of MnF3 into grams:
Mass of MnF3 = Moles of MnF3 * molar mass of MnF3
Finally, multiply the theoretical yield with the percentage yield (56%) to obtain the expected yield in grams.
Expected yield of MnF3 = Theoretical yield * Percentage yield
By following these steps, you can calculate the expected grams of MnF3 in the given reaction.
.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
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