Freon-12 is formed as follows:
3 CCl4 + 2 SbF3 -> 3 CCl2F2 + 2 SbCl3
150 g tetra chloromethane is combined with 100 g antimony (III) fluoride to give Freon-12 (CCl2F2).
Identify the limiting and excess reagents, and calculate how many grams of Freon-12 can be formed.
mols CCl4 = grams/molar mass = ?
mols SbF3 = grams/molar mass = ?
Convert mols CCl4 to mols SbF3. If the SbF3 mols is greater than CCl4 is the limiting reagent. If not SbF3 is the LR.
Post your work if you get stuck.
I got SbF3 as the limiting reagent.
What is the excess reagents ?
Well, let's get our chemistry clown wig on and solve this problem with a smile!
To find the limiting reagent, we need to compare the number of moles of each reactant with their respective stoichiometric coefficients.
First, let's calculate the number of moles for each compound. Remember the molar mass of CCl4 is 153.82 g/mol, SbF3 is 178.75 g/mol, and CCl2F2 is 120.91 g/mol.
150 g CCl4 x (1 mol/153.82 g) = 0.975 mol CCl4
100 g SbF3 x (1 mol/178.75 g) = 0.559 mol SbF3
Now, let's compare the moles of each reactant to their stoichiometric coefficients. According to the balanced equation, the ratio of CCl4 to SbF3 is 3:2.
So, we have:
0.975 mol CCl4 x (2 mol SbF3/3 mol CCl4) = 0.65 mol SbF3
Comparing this value to the actual moles of SbF3, we can see that SbF3 is the limiting reagent because we have less than the required 0.65 mol.
Now, let's calculate the theoretical yield of Freon-12 (CCl2F2) using SbF3 as the limiting reagent. According to the balanced equation, the stoichiometric ratio of SbF3 to CCl2F2 is 2:3.
0.559 mol SbF3 x (3 mol CCl2F2/2 mol SbF3) x (120.91 g/1 mol) = 100.15 g CCl2F2
So, the maximum yield of Freon-12 (CCl2F2) is 100.15 grams.
Now, as the chemistry clown, I must caution you to handle reagents and calculations safely without clowning around too much. Safety first!
I hope this colorful answer brings a smile to your face!
To identify the limiting and excess reagents, we need to compare the number of moles of each reactant to the stoichiometric ratio given by the balanced equation.
Let's calculate the number of moles for both reactants:
The molar mass of CCl4 (tetra chloromethane) is:
C = 12.01 g/mol * 1 = 12.01 g/mol
Cl = 35.45 g/mol * 4 = 141.8 g/mol
Total molar mass = 12.01 g/mol + 141.8 g/mol = 153.81 g/mol
Using the given mass of tetra chloromethane (CCl4):
Moles of CCl4 = mass / molar mass = 150 g / 153.81 g/mol
The molar mass of SbF3 (antimony (III) fluoride) is:
Sb = 121.76 g/mol * 1 = 121.76 g/mol
F = 18.998 g/mol * 3 = 56.994 g/mol
Total molar mass = 121.76 g/mol + 56.994 g/mol = 178.754 g/mol
Using the given mass of antimony (III) fluoride (SbF3):
Moles of SbF3 = mass / molar mass = 100 g / 178.754 g/mol
Now, using the balanced equation, we can determine the stoichiometric ratio:
3 CCl4 + 2 SbF3 -> 3 CCl2F2 + 2 SbCl3
From the equation, we can see that 3 moles of CCl4 react with 2 moles of SbF3 to form 3 moles of CCl2F2.
Let's calculate the moles of CCl2F2 that can be formed from each reactant:
Moles of CCl2F2 from CCl4 = Moles of CCl4 * (3 moles CCl2F2 / 3 moles CCl4)
Moles of CCl2F2 from SbF3 = Moles of SbF3 * (3 moles CCl2F2 / 2 moles SbF3)
Now, compare these values to determine the limiting and excess reagents:
The reactant that produces fewer moles of CCl2F2 is the limiting reagent.
The reactant that produces more moles of CCl2F2 is the excess reagent.
Let's calculate the moles of CCl2F2 formed from each reactant:
Moles of CCl2F2 from CCl4 = (150 g / 153.81 g/mol) * (3 moles CCl2F2 / 3 moles CCl4)
Moles of CCl2F2 from SbF3 = (100 g / 178.754 g/mol) * (3 moles CCl2F2 / 2 moles SbF3)
Now, compare these values to determine the limiting and excess reagents:
The limiting reagent: The reactant that produces fewer moles of CCl2F2.
The excess reagent: The reactant that produces more moles of CCl2F2.
Next, let's calculate the theoretical yield of Freon-12 (CCl2F2) formed using the limiting reagent:
From the balanced equation, we can see that the stoichiometric ratio between CCl2F2 and CCl4 is 3:3, meaning 3 moles of CCl4 react with 3 moles of CCl2F2.
Moles of CCl4 reacted = Moles of CCl2F2 formed
Moles of CCl4 reacted = Moles of CCl2F2 from CCl4
Now, calculate the mass of CCl2F2 using the molar mass:
Mass of CCl2F2 = Moles of CCl2F2 * Molar mass of CCl2F2
This will give you the maximum quantity (in grams) of Freon-12 (CCl2F2) that can be formed in this reaction.