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 51%, how many grams of MnF3 can be expected if 40.0 grams of each reactant is used in an experiment?
You have to convert each 40 grams to moles of reactants.
Then, using the mole ratio 2:13, determine which is the limiting reactant. I assume you can do that. Then, say the limiting reactant is F2 (It probably wont be the real limiting reactant here), and you had XXX moles of it to react. Then moles of MnF3 you get will be .50*2/13*mole of F2 you really had.
Oh yes, then convert those moles of MnF3 to grams.
To find the expected grams of MnF3 produced, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in a chemical reaction, thereby limiting the amount of product that can be formed.
Step 1: Calculate the molar mass of each reactant.
Molar mass of MnI2 = 317.808 g/mol
Molar mass of F2 = 38.003 g/mol
Step 2: Convert the mass of each reactant to moles.
Moles of MnI2 = (mass of MnI2) / (molar mass of MnI2)
Moles of MnI2 = 40.0 g / 317.808 g/mol = 0.126 moles
Moles of F2 = (mass of F2) / (molar mass of F2)
Moles of F2 = 40.0 g / 38.003 g/mol = 1.052 moles
Step 3: Use the stoichiometric coefficients to calculate the moles of MnF3 formed.
From the balanced equation, we know that 2 moles of MnI2 react with 13 moles of F2 to produce 2 moles of MnF3.
Moles of MnF3 formed = (moles of limiting reactant) * (moles of MnF3 / moles of limiting reactant)
Moles of MnF3 formed = 0.126 moles * (2 moles MnF3 / 2 moles MnI2) = 0.126 moles
Step 4: Calculate the theoretical yield of MnF3.
The theoretical yield is the maximum amount of product that can be obtained under ideal conditions.
Theoretical yield of MnF3 = (moles of MnF3 formed) * (molar mass of MnF3)
Theoretical yield of MnF3 = 0.126 moles * (150.926 g/mol) = 19.000 g
Step 5: Calculate the expected yield of MnF3 based on the percentage yield.
The expected yield is the actual amount of product obtained in the given experiment, taking into account the percentage yield.
Expected yield of MnF3 = (percentage yield / 100) * (theoretical yield)
Expected yield of MnF3 = 51% / 100 * 19.000 g = 9.690 g
Therefore, approximately 9.690 grams of MnF3 can be expected if 40.0 grams of each reactant is used in the experiment.