The reaction of 5.7 grams of fluorine with
excess chlorine produced 7.6 grams of ClF3.
What percent yield of ClF3 was obtained?
Answer in units of %.
3F2 + Cl2 ==> 2ClF3
mols F2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols F2 to mols ClF3. That will be mols F2 x (2/3) = ?
Then convert mols ClF3 to grams. g = mols x molar mass = ? This is the theoretical yield (TY).
You know from the problem that the actual yield (AY) is 7.6 g.
%yield = (AY/TY)*100 = ?
82.6%
To find the percent yield of ClF3, we need to compare the actual yield (7.6 grams) with the theoretical yield. The theoretical yield is the amount of ClF3 that would be produced if the reaction went to completion.
First, let's calculate the molar mass of ClF3:
Molar mass ClF3 = (1 * molar mass Cl) + (3 * molar mass F)
= (1 * 35.45 g/mol) + (3 * 18.998 g/mol)
= 35.45 g/mol + 56.994 g/mol
= 92.444 g/mol
Next, let's calculate the theoretical yield of ClF3:
Theoretical yield of ClF3 = (mass of F2 used / molar mass F2) * (molar mass ClF3 / 2)
= (5.7 g / 37.996 g/mol) * (92.444 g/mol / 2)
= (0.15003 mol) * (46.222 g/mol)
= 6.931 g
Now, we can calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100%
= (7.6 g / 6.931 g) * 100%
= 109.77%
Therefore, the percent yield of ClF3 obtained in this reaction is 109.77%.
To calculate the percent yield of ClF3, we need to compare the actual yield (7.6 grams) with the theoretical yield. The theoretical yield is the amount of ClF3 that would be obtained if the reaction went to completion, based on the balanced chemical equation.
First, we need to determine the limiting reactant. The limiting reactant is the one that gets completely consumed and determines the amount of product that can be formed. In this case, we know that fluorine (F2) was the limiting reactant since it reacted with excess chlorine (Cl2).
To determine the moles of fluorine used, we divide its mass by its molar mass:
moles of F2 = mass of F2 / molar mass of F2
moles of F2 = 5.7 g / 38.0 g/mol (molar mass of F2)
moles of F2 = 0.15 mol
From the balanced chemical equation, we can see that for every 1 mole of F2, we should obtain 1 mole of ClF3. Therefore, the theoretical yield of ClF3 can be calculated using the moles of F2:
moles of ClF3 = moles of F2 = 0.15 mol
Now, we need to convert the moles of ClF3 to grams:
mass of ClF3 = moles of ClF3 × molar mass of ClF3
mass of ClF3 = 0.15 mol × 92.44 g/mol (molar mass of ClF3)
mass of ClF3 = 13.866 g
The percent yield can be calculated using the formula:
percent yield = (actual yield / theoretical yield) × 100
percent yield = (7.6 g / 13.866 g) × 100
percent yield = 54.84%
Therefore, the percent yield of ClF3 obtained is 54.84%.