Asked by byke
Gaseous iodine pentafluoride, IF5, can be prepared by the reaction of solid iodine and
gaseous fluorine:
• I2 (s) + 5F2 (g) → 2IF5 (g).
• A 5.00 L flask is filled with 10.0 g I2 and excess F2 with the reaction proceeds to
completion completely consumed. After the reaction is complete the temperature in
the flask is 125°C.
i) What is the pressure of IF5 (g) in the flask ?
– ii) How many molecules of F2 reacted with I2 (s) ?
– iii) What mass of IF5 (g) is produce.
gaseous fluorine:
• I2 (s) + 5F2 (g) → 2IF5 (g).
• A 5.00 L flask is filled with 10.0 g I2 and excess F2 with the reaction proceeds to
completion completely consumed. After the reaction is complete the temperature in
the flask is 125°C.
i) What is the pressure of IF5 (g) in the flask ?
– ii) How many molecules of F2 reacted with I2 (s) ?
– iii) What mass of IF5 (g) is produce.
Answers
Answered by
DrBob222
I THINK the problem says there is noting present at the end of the reaction except IF5.
mols I2 = grams/molar mass = ?
Convert mols I2 to mols IF5 using the coefficients in the balanced equation.
Use PV = nRT and solve for P in atm.
Use mols I2 and convert to mols F2, then multiply by 6.02E23 to find molecules F2.
g IF5 = mols IF5 x molar mass IF5.
mols I2 = grams/molar mass = ?
Convert mols I2 to mols IF5 using the coefficients in the balanced equation.
Use PV = nRT and solve for P in atm.
Use mols I2 and convert to mols F2, then multiply by 6.02E23 to find molecules F2.
g IF5 = mols IF5 x molar mass IF5.
Answered by
byke
thanks Dr.Bob222 you're the best
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