How many gram of fluorine must be reacted with 12.0g of MnI2 to produce a 75% yield in this equation: 2MnI2+13F2=2MnF3+4IF5

• 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

• Moles of MnI2 = 12.0 g / 308.746 g/mol = 0.038864 mol MnI2

• 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

• Mass of F2 (theoretical) = 0.252616 mol * 37.996 g/mol = 9.6004 g F2

• 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.