Question
IF5 can be prepared by the following reaction: I2+5F2--->2IF5. A 5.00liter flask containing 10.0g I2 is filledwith 10.0g of F2, and the reaction proceeds until one of thereagents is completely consumed. After the reaction is complete,the temperature in the flask is 125 C. calculate the partialpressure of IF5 in the flask. I2 molar mass=253.8g/mol F2 molarmass=38.00g/mol IF5 molar mass=221.9g/mol IF5 can be prepared by the following reaction: I2+5F2--->2IF5. A 5.00liter flask containing 10.0g I2 is filledwith 10.0g of F2, and the reaction proceeds until one of thereagents is completely consumed. After the reaction is complete,the temperature in the flask is 125 C. calculate the partialpressure of IF5 in the flask. I2 molar mass=253.8g/mol F2 molarmass=38.00g/mol IF5 molar mass=221.9g/mol
Explanation / Answer
I2 +5F2--->2IF5 253.8 g of I 2 reacts with 5*38 g of F 2 produces 2 * 221.9 g of IF 5 10 g of I 2 requires X g of F 2 X = [ 5 * 38 * 10 ] / 253.8 = 7.4862 g Therefore the mass of F 2 unreacted = 10 - 7.4862 = 2.5138g 253.8 g of I 2 produces 443.8 g of IF5 10 g of I2 produces Y g of IFO So, Y = [443.8 * 10 ] / 253.8 =17.4862 g Given Volume V = 5 L temperature T = 125 oC = 125 + 273 = 398 K Molecular weight of IF 5 is M = 221.9 g Weight of IF 5 produced w = 17.4862 g gas constant R =0.0821 L atm / mol-K from gas law PV = nRT = ( w / M ) RT Therefore partial pressure of IF 5 is P = ( wRT /MV ) = 0.51498 atm = 253.8 g of I 2 produces 443.8 g of IF5 10 g of I2 produces Y g of IFO So, Y = [443.8 * 10 ] / 253.8 =17.4862 g Given Volume V = 5 L temperature T = 125 oC = 125 + 273 = 398 K Molecular weight of IF 5 is M = 221.9 g Weight of IF 5 produced w = 17.4862 g gas constant R =0.0821 L atm / mol-K from gas law PV = nRT = ( w / M ) RT Therefore partial pressure of IF 5 is P = ( wRT /MV ) = 0.51498 atm =