The reaction of 5.3 grams of fluorine with

excess chlorine produced 4.6 grams of ClF3. What percent yield of ClF3 was obtained?
Answer in units of %

To determine the percent yield of ClF3, we need to compare the actual yield (4.6 grams) to the theoretical yield. The theoretical yield is the maximum amount of product that could be formed based on the balanced chemical equation and the amount of limiting reagent used.

First, we need to identify the limiting reagent. The limiting reagent is the reactant that is completely consumed, thereby limiting the amount of product that can be formed. To do this, we calculate the moles of fluorine and chlorine in the reaction using their respective molar masses:

Molar mass of F2 = 2 * (atomic mass of fluorine) = 2 * 19.00 g/mol = 38.00 g/mol
Molar mass of Cl2 = 2 * (atomic mass of chlorine) = 2 * 35.45 g/mol = 70.90 g/mol

Next, we calculate the moles of fluorine (F2) and chlorine (Cl2) using their respective masses:

Moles of F2 = mass of F2 / molar mass of F2 = 5.3 g / 38.00 g/mol
Moles of Cl2 = excess (since it is in excess)

The balanced chemical equation for the reaction is:

2 F2 + 3 Cl2 → 2 ClF3

The stoichiometric ratio between F2 and ClF3 is 2:2, which means that for every 2 moles of F2 used, we should ideally produce 2 moles of ClF3.

Therefore, the moles of ClF3 that could be formed can be calculated using the moles of F2:

Moles of ClF3 = moles of F2 * (moles of ClF3 / moles of F2)
= moles of F2 * (2 mole ClF3 / 2 mole F2)
= moles of F2

Since we know the moles of F2 produced, we can convert it to grams using the molar mass of ClF3:

Theoretical yield of ClF3 = moles of ClF3 * molar mass of ClF3 = moles of F2 * molar mass of ClF3

Now we can calculate the percent yield using the formula:

Percent yield = (actual yield / theoretical yield) * 100

Substituting the given values:

Percent yield = (4.6 g / theoretical yield) * 100

This calculation depends on the amount of excess chlorine used, which is not provided in the question. If the amount of excess chlorine is specified, we can proceed with the calculation. Otherwise, we would need additional information to determine the percent yield.

Yield = practical/theoretical

Given:
Practical mass: Pm{ClF3} = 4.6g
Reactant mass: m{F2} = 5.3g
The equation: 3 F2 + Cl2 = 2 ClF3

Lookup the Molecular Weights:
w{F2} = 37.99680650 ± 0.00000002 g/mol
w{ClF3} = 92.4484 ± 0.0001 g/mol

So theoretically, the mass of chlorine triflouride is:
Tm{ClF3} = (2/3) m{F2}w{ClF3}/w{F2}

Therefore:
Yield = Pm{ClF3}/Tm{ClF3}
= (3/2) Pm{ClF3} w{F2} /(m{F2} w{ClF3})
= (3/2)*4.6*37.99680650/(5.3*92.4484)
= 53.5%