Final answer:
The theoretical yield of iron(III) chloride is 7.61 grams based on stoichiometric calculations. Using the actual yield of 5.37 grams, the percent yield is calculated to be 70.57%, which does not match any provided options.
Step-by-step explanation:
To determine the theoretical yield of iron(III) chloride, we need to perform stoichiometric calculations based on the balanced chemical reaction between iron and chlorine. First, we convert the mass of chlorine used to moles, then use the molar ratio from the balanced equation to find the moles of iron(III) chloride that should form if all the chlorine reacts completely. Finally, we convert the moles of iron(III) chloride to mass to find the theoretical yield.
The formula weight of Cl2 is approximately 70.90 grams/mole (35.45 × 2), and the formula weight of FeCl3 is approximately 162.20 grams/mole (55.85 + 35.45 × 3). Given that 4.99 grams of chlorine gas is used, we calculate the moles of Cl2 as follows:
Moles of Cl2 = 4.99 g / 70.90 g/mol = 0.0704 moles
Using the stoichiometry of the reaction, where 3 moles of Cl2 react with iron to produce 2 moles of FeCl3, we calculate the moles of FeCl3:
Moles of FeCl3 = 0.0704 moles Cl2 × (2 moles FeCl3 / 3 moles Cl2) = 0.0469 moles FeCl3
Now, converting moles of FeCl3 to mass:
Mass of FeCl3 = 0.0469 moles × 162.20 g/mol = 7.61 grams
Therefore, the theoretical yield of FeCl3 is 7.61 grams, which corresponds to option A) 6.12 grams, the closest answer provided.
To find the percent yield, we compare the actual yield to the theoretical yield:
Percent Yield = (Actual Yield / Theoretical Yield) × 100% = (5.37 g / 7.61 g) × 100% = 70.57%
However, there is no option given that matches 70.57%, indicating a possible error in the question or the provided options.