A gas is collected by water displacement so that its partial pressure is 72.92 kPa. The total pressure of the gas over water is 75.02 kPa. What is the temperature at which the gas was collected?

18.5 degree Celsius

Explanation:

We are given that

Partial pressure of gas==72.92 KPa

Total pressure gas =P=75.02 KPa

We have to find the temperature at which the gas was collected.

We know that

Total pressure,P=

Water vapor pressure=

101.3 KPa=760 mm Hg

Temperature when water vapor pressure is 15.75 mm Hg=18.5 degree Celsius

The temperature at which gas collected=18.5 degree Celsius

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"