The reaction of 9.6 grams of fluorine with excess chlorine produced 3.2 grams of clf3. what percent yield of clf3 was obtained?

First, you have to know the reaction because stoichiometric coefficients are essential for our calculations later on. We know the reactants to be fluorine and chlorine, and the product is ClF₃. Now, you should note that Fluorine and Chlorine are 2 of the 5 diatomic gases that occur in nature. So, in chemical formula, they are written as Cl₂ and F₂. After balancing the equation, the complete reaction would be:

3 F₂ + Cl₂ ⇒ 2 ClF₃

Next, you should distinguish which of the reactants is limiting. As mentioned in the problem, it is Fluorine. Therefore, we base our calculations on the initial amount of Fluorine. We compute for the theoretical yield. Fluorine has a molar mass of 19 g/mol, while ClF₃ has a molar mass of 92.45 g/mol.

Theoretical yield = 9.6 g F₂ * (1 mol F₂/19 g) * (2 mol ClF₃/ 3 mol F₂) * (92.45 g/mol)
Theoretical yield = 31.14 g

The actual yield is 3.2 grams. To compute for percent yield, the formula is:

Percent yield = (Actual yield/Theoretical yield)*100
Percent yield = (3.2 g/31.14 g)*100
Percent yield = 10.28%