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 %
1. Write and balance the equation.
2. Convert 8.7g F2 to mols. mols = grams/molar mass.
3. Using the coefficients in the balanced equation, convert mols F2 to mols of the product.
4. Convert mols product to grams. g = mols x molar mass. The is the theoretical yield (TE).\
5. %yield = (actual yield/TE)*100 = ?
3.9 g is the actual yield.
To calculate the percent yield of a reaction, you need to compare the actual yield of the product with the theoretical yield of the product. The percent yield is then calculated using the following formula:
Percent yield = (Actual Yield / Theoretical Yield) * 100
In this case, we are given the mass of the reactant (fluorine) and the mass of the product (ClF3). However, we are not provided with the theoretical yield. To find the theoretical yield, we need to first identify the limiting reactant.
The limiting reactant is the reactant that is completely consumed in a chemical reaction, thereby determining the maximum amount of product that can be formed. In this case, we are told that chlorine is in excess, which means that fluorine is the limiting reactant.
To determine the theoretical yield of ClF3, we need to calculate the amount of ClF3 that can be formed from the given mass of fluorine. To do this, we first need to find the molar mass of fluorine (F) and ClF3.
The molar mass of F = 19.0 g/mol
The molar mass of ClF3 = 83.45 g/mol (35.45 g/mol for Cl + 3 * 19.0 g/mol for F)
Now we can calculate the moles of F:
Moles of F = mass / molar mass = 8.7 g / 19.0 g/mol = 0.458 mol
Since the stoichiometry of the balanced chemical equation is 1:1 (1 mol of F reacts with 1 mol of ClF3), the moles of ClF3 that can be formed will also be 0.458 mol.
Now we can find the theoretical yield of ClF3:
Theoretical Yield = moles of ClF3 * molar mass of ClF3 = 0.458 mol * 83.45 g/mol = 38.245 g
Finally, we can calculate the percent yield:
Percent yield = (Actual Yield / Theoretical Yield) * 100
= (3.9 g / 38.245 g) * 100
= 10.19%
Therefore, the percent yield of ClF3 in this reaction is approximately 10.19%.