This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
To determine the mass of fluorine required, we'll follow these steps: 1. Calculate the molar mass of MnI2 and F2. Molar mass of MnI2 = 54.938 (Mn) + 2 126.904 (I) = 54.938 + 253.808 = 308.746 g/mol Molar mass of F2 = 2 18.998 (F) = 37.996 g/mol 2. Convert the given mass of MnI2 to moles. Moles of MnI2 = 12.0 g / 308.746 g/mol = 0.038864 mol MnI2 3. Use the stoichiometry of the balanced equation to find the theoretical moles of F2 needed for a 100% yield. From the equation (2MnI2 + 13F2 = 2MnF3 + 4IF5), 2 moles of MnI2 react with 13 moles of F2. Moles of F2 (theoretical) = 0.038864 mol MnI2 (13 mol F2 / 2 mol MnI2) = 0.252616 mol F2 4. Convert the theoretical moles of F2 to grams. Mass of F2 (theoretical) = 0.252616 mol 37.996 g/mol = 9.6004 g F2 5. Adjust for the 75% yield. If the reaction has a 75% yield, it means that to achieve the theoretical amount of product (which is based on the 12.0g of MnI2), you need to supply more reactant (F2) to compensate for the inefficiency. Mass of F2 required = (Theoretical mass of F2) / (Percent yield as a decimal) Mass of F2 required = 9.6004 g / 0.75 = 12.8005 g Rounding to three significant figures (based on 12.0g MnI2), 12.8 grams of fluorine must be reacted. Answer: 12.8 grams of fluorine must be reacted with 12.0g of MnI2 to produce a 75% yield.