The reaction of 8 grams of fluorine with excess
chlorine produced 6.4 grams of ClF3. What
percent yield of ClF3 was obtained?
Answer in units of %
To find the percent yield of ClF3, we need to compare the actual yield (the amount of ClF3 that was produced in the reaction) to the theoretical yield (the amount of ClF3 that would be produced if the reaction went to completion).
First, we need to calculate the theoretical yield of ClF3. We can do this by using the balanced chemical equation for the reaction between fluorine (F2) and chlorine (Cl2) to produce ClF3:
F2 + Cl2 → 2ClF3
The molar mass of ClF3 is 92.45 g/mol (1 Cl atom + 3 F atoms). Using the molar mass of ClF3, we can calculate the moles of ClF3 produced from the given mass of ClF3:
moles of ClF3 = mass of ClF3 / molar mass of ClF3
moles of ClF3 = 6.4 g / 92.45 g/mol
Next, we need to calculate the moles of F2 that reacted. Since the reaction of F2 with Cl2 is in a 1:1 ratio, the moles of ClF3 produced is equal to the moles of F2 reacted. We can use the molar mass of F2 (38.00 g/mol) to calculate the moles of F2:
moles of F2 = moles of ClF3
moles of F2 = 6.4 g / 92.45 g/mol
Now we can calculate the mass of F2 that reacted:
mass of F2 = moles of F2 * molar mass of F2
mass of F2 = (6.4 g / 92.45 g/mol) * 38.00 g/mol
Finally, we can calculate the percent yield:
percent yield = (actual yield / theoretical yield) * 100%
percent yield = (6.4 g / (6.4 g / 92.45 g/mol) * 38.00 g/mol) * 100%
Simplifying the equation:
percent yield = (6.4 / (6.4 / 92.45) * 38) * 100%
percent yield = (6.4 / 1) * 100%
percent yield = 640%
Therefore, the percent yield of ClF3 is 640%.
To calculate the percent yield of ClF3, we need to compare the actual yield (6.4 grams) with the theoretical yield. The theoretical yield is the maximum amount of product that can be obtained based on stoichiometry.
First, we need to determine the balanced chemical equation for the reaction between fluorine (F2) and chlorine (Cl2) to form ClF3:
F2 + Cl2 → 2ClF3
From the balanced equation, we can see that 1 mole of F2 reacts with 1 mole of Cl2 to produce 2 moles of ClF3.
To find the theoretical yield of ClF3, we need to calculate the number of moles of F2 used in the reaction.
Given that the mass of F2 is 8 grams and its molar mass is 19.0 g/mol, we can use the formula:
moles of F2 = mass of F2 / molar mass of F2
moles of F2 = 8 g / 19.0 g/mol = 0.4211 mol
Since 1 mole of F2 reacts with 2 moles of ClF3, we can calculate the theoretical yield of ClF3 using the coefficient ratio:
moles of ClF3 = 2 * moles of F2
moles of ClF3 = 2 * 0.4211 mol = 0.8422 mol
Now, we can calculate the theoretical yield of ClF3 in grams using the molar mass of ClF3:
mass of ClF3 = moles of ClF3 * molar mass of ClF3
mass of ClF3 = 0.8422 mol * (35.45 g/mol + 18.99 g/mol) = 0.8422 mol * 54.44 g/mol = 45.92 g
The theoretical yield of ClF3 is 45.92 grams.
Finally, we can calculate the percent yield using the formula:
percent yield = (actual yield / theoretical yield) * 100
percent yield = (6.4 g / 45.92 g) * 100
percent yield = 13.93%
Therefore, the percent yield of ClF3 obtained in this reaction is 13.93%.
Cl2 + 3F2 ==> 2ClF3
mols F2 = g/molar mass = ?
Using the coefficients in the balanced equation, convert mols F2 to mols Cl
F3.
Now convert mols ClF3 to grams. g = mols x molar mass = ? This is the theoretical yield (TY) assuming 100% of the product is formed.The actual yield (AY) in the problem is 6.4 g.
Then % yield = [(AY)/(TY)]*100
Post your work if you get stuck.