look up the vapor pressure of H2O at SATP. SUbtract that from SP, that is H2 pressure.
Now, knowing H2 pressure in the jar, and volume, and temp, calculate the moles of H2 collected (PV=nRT)
Under SATP, magnesium was reacted with acid and 500mL of hydrogen gas was collected over water
a) What is the vapour pressure of water?
b) What is the pressure of hydrogen gas?
c) What amount of hydrogen was produced?
5 answers
a. Don't you have a table in the book (or look it up on the web) to find the vapor pressure.
b. Ptotal = pH2 + pH2O. Ptotal is 1 atm and you know pH2O, solve for pH2.
c. Use PV = nRT. P from above, V is given, you know R and T is 25C (converted to K).
Post your work if you get stuck.
b. Ptotal = pH2 + pH2O. Ptotal is 1 atm and you know pH2O, solve for pH2.
c. Use PV = nRT. P from above, V is given, you know R and T is 25C (converted to K).
Post your work if you get stuck.
This is my answer below, it is close to the texbook answer but not exact.
a) V pressure of H2O = 25 degrees celcuis
= 3.17kPa H2O
(This was in my texbook by the way, I just didn't see it)
b) Pt=pH2+pH2O pt= 1 atm= 101.3kPa
101.3kPa =pH2 + 3.17kPa
101.3kPa - 3.17kPa = pH2
pH2= 98.13kPa (The textbook is 96.8kPa)
c) PV= nRT
98.13(0.55l) divided by 8.314kpa.l/mol.K = 0.0217
n= 0.0217
a) V pressure of H2O = 25 degrees celcuis
= 3.17kPa H2O
(This was in my texbook by the way, I just didn't see it)
b) Pt=pH2+pH2O pt= 1 atm= 101.3kPa
101.3kPa =pH2 + 3.17kPa
101.3kPa - 3.17kPa = pH2
pH2= 98.13kPa (The textbook is 96.8kPa)
c) PV= nRT
98.13(0.55l) divided by 8.314kpa.l/mol.K = 0.0217
n= 0.0217
I would have worked the problem almost the same except I would have used101.325 since you used 3.17 which gives a slightly different answer of 98.15. That will change the moles slightly, too. However, I think I know how the book obtained 96.8 kPa. Standard pressure is often used as 1 bar or 100 kPa and 100-3.17 = 96.8 kPa.
I agree with DrBob. When the SATP was being debated, standard pressure was often debated to be 100kPa as "convenient"