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.).   :-)

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

i just learnt this recently but i need to look through my notes again