Question

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

Answer 1

Approx. `200` `L` dioxygen gas.

Explanation:

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"`).