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

Apr 9, 2016

Approx. $200$ $L$ dioxygen gas.

#### Explanation:

The molar quantity of $225$ $m L$ of water at 27""^@ C is $\frac{225 \cdot m L \times 1.00 \cdot g \cdot m {L}^{-} 1}{18.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $12.5 \cdot m o l$.

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:

${H}_{2} O \left(g\right) \rightarrow {H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right)$

Stoichiometry clearly shows that $6.25$ $m o l$ dioxygen gas would result if $12.5$ $m o l$ 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:

$V = \frac{n R T}{P}$ $=$ $\frac{6.25 \cdot \cancel{m o l} \times 0.0821 \cdot L \cdot \cancel{a t m} \cdot \cancel{{K}^{-} 1 \cdot m o {l}^{-} 1} \times 300 \cdot \cancel{K}}{0.763 \cdot \cancel{a t m}}$ $=$ ??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: $R = 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1$ 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. $1 \cdot a t m \equiv 760 \text{ mm Hg}$).