The reaction of 8.7 grams of fluorine with
excess chlorine produced 3.9 grams of ClF3.
What percent yield of ClF3 was obtained?
Answer in units of %
I thought I did this for you yesterday.
Here is a worked example of a limiting reagent problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
To calculate the percent yield of ClF3, we need to compare the actual yield (3.9 grams) with the theoretical yield. The theoretical yield is the amount of ClF3 that would be produced if the reaction occurred with 100% efficiency.
To find the theoretical yield, we need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
In this case, we know that there is an excess of chlorine, so the limiting reagent is fluorine. To find the amount of ClF3 that would be produced from the limiting reagent, we can use the stoichiometry of the balanced chemical equation.
The balanced chemical equation for the reaction between fluorine (F2) and chlorine (Cl2) to form ClF3 is:
F2 + Cl2 -> 2ClF3
From the equation, we can see that 2 moles of ClF3 are produced for every 1 mole of F2 consumed.
First, we need to convert the mass of F2 (8.7 grams) to moles. We can use the molar mass of F2 to do this. The molar mass of F2 is approximately 38.00 g/mol (19.00 g/mol per F atom), so:
moles of F2 = 8.7 grams / 38.00 g/mol ≈ 0.229 moles
Now, using the stoichiometry of the balanced equation, we can determine how many moles of ClF3 would be produced:
moles of ClF3 = 2 moles of ClF3 / 1 mole of F2 * 0.229 moles of F2 = 0.458 moles
Finally, we can convert the moles of ClF3 to grams:
grams of ClF3 = moles of ClF3 * molar mass of ClF3
= 0.458 moles * (35.45 g/mol + 19.00 g/mol + 19.00 g/mol) ≈ 26.830 grams
Now that we have the theoretical yield, we can calculate the percent yield:
percent yield = (actual yield / theoretical yield) * 100%
= (3.9 grams / 26.830 grams) * 100%
≈ 14.52%
Therefore, the percent yield of ClF3 obtained in this reaction is approximately 14.52%.