Manganese trifluoride, MnF3,can be prepared by the following reaction.
2 MnI2 (s) + 13 F2 (g) ---> 2 MnF3(s) + 4 IF5(l)
What is the minimum number of grams of F2 that must be used to react with 12.0 g of MnI2 if the overall yield of MnF3 is no more than 75% of theory?
To find the minimum number of grams of F2 required to react with 12.0 g of MnI2, we need to calculate the stoichiometry of the reaction.
Step 1: Calculate the molar mass of MnI2 (manganese iodide):
Molar mass of Mn = 54.94 g/mol
Molar mass of I = 126.9 g/mol
Molar mass of MnI2 = (54.94 g/mol) + 2 * (126.9 g/mol) = 308.74 g/mol
Step 2: Calculate the number of moles of MnI2:
Number of moles = Mass / Molar mass = 12.0 g / (308.74 g/mol) = 0.0389 mol
Step 3: Determine the stoichiometry of the reaction:
From the balanced equation, we can see that 2 moles of MnI2 react with 13 moles of F2 to produce 2 moles of MnF3.
Step 4: Calculate the theoretical yield of MnF3:
Theoretical yield = (0.0389 mol MnI2) * (2 mol MnF3 / 2 mol MnI2) = 0.0389 mol MnF3
Step 5: Calculate the maximum possible yield of MnF3:
Maximum possible yield = 0.75 * Theoretical yield = 0.75 * 0.0389 mol = 0.0292 mol
Step 6: Calculate the number of moles of F2 required:
From the balanced equation, we can see that 13 moles of F2 react with 2 moles of MnF3.
Number of moles of F2 = (0.0292 mol MnF3) * (13 mol F2 / 2 mol MnF3) = 0.1907 mol
Step 7: Calculate the mass of F2 required:
Mass = Number of moles * Molar mass = 0.1907 mol * (38.0 g/mol, molar mass of F2) = 7.2394 g
Therefore, the minimum number of grams of F2 that must be used to react with 12.0 g of MnI2 is approximately 7.24 g.
To determine the minimum number of grams of F2 required to react with 12.0 g of MnI2, we need to consider the stoichiometry of the reaction and the yield of MnF3.
1. Calculate the molar mass of MnI2:
Molar mass of Mn = 55 g/mol
Molar mass of I = 127 g/mol
Molar mass of MnI2 = (2*55) + (2*127) = 281 g/mol
2. Convert the mass of MnI2 to moles:
Moles of MnI2 = 12.0 g / 281 g/mol = 0.0427 mol
3. Use the balanced equation to determine the stoichiometry between MnI2 and F2:
According to the balanced equation, 2 moles of MnI2 react with 13 moles of F2 to produce 2 moles of MnF3.
Therefore, the mole ratio between MnI2 and F2 is:
2 mol MnI2 : 13 mol F2
4. Determine the lower limit of moles of F2 required:
Moles of F2 = (0.0427 mol MnI2) * (13 mol F2 / 2 mol MnI2) = 0.2779 mol F2
5. Calculate the minimum grams of F2 required:
Mass of F2 = (0.2779 mol F2) * (38.003 g/mol) = 10.558 g F2
6. Consider the yield of MnF3:
If the overall yield of MnF3 is no more than 75% of the theoretical yield, we need to adjust the calculated mass of F2 accordingly.
Minimum grams of F2 = (10.558 g F2) / (0.75) = 14.077 g F2
Therefore, the minimum number of grams of F2 that must be used to react with 12.0 g of MnI2 is approximately 14.077 grams.
12.0g MnI2/0.75 = ? g MnI2 to start accounting for the 75% yield = approx 16
mols MnI2 = grams/molar mass = 16/about 300 = ?
Convert to mols F2. That's
? mols MnI2 x (13mols F2/2 mols MnI2) = ? x 13/2 = x mols MnI2.
Convert to grams. g = mols x molar mass