The reaction of 5.1 grams of fluorine with excess chlorine produced 4.5 grams of ClF3. What percent yield of ClF3 was obtained? Answer in unites of %

To determine the percent yield of ClF3, we need to compare the actual yield of ClF3 to the theoretical yield. The theoretical yield is the maximum amount of ClF3 that can be obtained based on the balanced chemical equation.

First, let's write the balanced chemical equation for the reaction between fluorine (F2) and chlorine (Cl2) to form ClF3:
F2 + 3Cl2 → 2ClF3

From the balanced equation, we can see that 1 mole of F2 reacts with 3 moles of Cl2 to produce 2 moles of ClF3.

1 mole of F2 is approximately 38 grams (based on the molar mass of F2), so 5.1 grams of F2 is equal to:
5.1 g / 38 g/mol ≈ 0.134 moles of F2

Since F2 is in excess, it is not the limiting reactant. Therefore, we need to determine the theoretical yield of ClF3 based on the amount of Cl2.

From the equation, we know that 3 moles of Cl2 react with 2 moles of ClF3. Thus, the number of moles of ClF3 that can be produced from Cl2 is:
0.134 moles of F2 × (2 moles of ClF3 / 3 moles of Cl2) ≈ 0.089 moles of ClF3

To determine the theoretical mass of ClF3, we need to multiply the number of moles by the molar mass of ClF3. The molar mass of ClF3 can be calculated as follows:
35.45 g/mol (molar mass of Cl) + 18.998 g/mol (molar mass of F) × 3 = 117.442 g/mol

The theoretical mass of ClF3 is:
0.089 moles of ClF3 × 117.442 g/mol ≈ 10.465 grams of ClF3

Now, we can calculate the percent yield by comparing the actual yield (4.5 grams) to the theoretical yield (10.465 grams):
Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (4.5 g / 10.465 g) × 100 ≈ 43.03%

Therefore, the percent yield of ClF3 obtained in this reaction is approximately 43.03%.