How many liters of oxygen gas are produced by the complete decomposition of 225 mL of water?

The molar quantity of #225# #mL# of water at #27""^@ C# is #(225*mLxx1.00*g*mL^-1)/(18.01*g*mol^-1)# #=# #12.5*mol#.

Now the decomposition of a molar quantity of water results in the formation of a HALF molar quantity of dioxygen gas, as given by the following equation:

#H_2O(g) rarr H_2(g) + 1/2O_2(g)#

Stoichiometry clearly shows that #6.25# #mol# dioxygen gas would result if #12.5# #mol# water were decomposed.

We must now convert that molar quantity of gas to a volume in litres given the stated conditions: #P=0.763*atm; T =300K#.

We idealize the behaviour of water and use the ideal gas equation:

#V=(nRT)/P# #=# #(6.25*cancel(mol)xx0.0821*L*cancel(atm)*cancel(K^-1*mol^-1)xx300*cancelK)/(0.763*cancel(atm))# #=# #??L#

What is the volume of dihydrogen produced under these conditions?

The obvious difficulty in solving these sorts of problems is selection of the Gas Constant, #R#. Many such constants exist: #R=0.0821*L*atm*K^-1*mol^-1# is probably the one that is most useful to chemists in that it uses sensible units such as litres, and atmospheres, both of which can be easily measured by chemists (i.e. #1*atm-=760" mm Hg"#).