Calculate the volume occupied by 55.5 moles of water at STP.

Volume occupied by 55.5 moles of water gas at STP = 1240 Liters.

Explanation:

1st keep in mind that 'volume' in the context of problems like this apply only to the gas phase form of the substance. Water in solid form (ice) or liquid form (liquid water) do not apply. Volume in this case is referred to as 'molar volume' and is a gas occupying 22.4 liters at STP conditions (0°C, 1Atm).

So, if 1 molar volume of water gas (steam) occupies 22.4 Liters at STP, then 55.5 moles of water gas occupies 22.4 Liters/mole x 55.5 moles = 1243.2 Liters, or 1240 Liters (3 sig. figs.).   :-)

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.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.