In one city, a balloon with a volume of 6.0 L is filled with air at 101 kPa
pressure. The balloon is then taken to a second city at a much higher
altitude. At this second city, atmospheric pressure is only 91 kPa. If the
temperature is the same in both places, what will be the new volume of the
balloon?

The new volume of the balloon is 6.6 L

Explanation:

To calculate the new tempearture, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

where,

are initial pressure and volume.

are final pressure and volume.

We are given:

Putting values in above equation, we get:

Thus new volume of the balloon is 6.6 L

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.

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