For the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ \(K=2.4 \times 10^{-3}\) at a given temperature. At equilibrium in a 2.0-L container it is found that \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=1.1 \times 10^{-1} M\) and \(\left[\mathrm{H}_{2}(g)\right]=1.9 \times 10^{-2} M .\) Calculate the moles of \(\mathrm{O}_{2}(g)\) present under these conditions.

Understanding the equilibrium constant expression is essential for anyone studying chemical equilibrium. The equilibrium constant, represented as K, quantifies the concentration ratio of products to reactants at equilibrium for a reversible reaction. It’s important to note that in the expression, gaseous or aqueous species are represented by their molar concentrations (M), while solids and liquids are omitted as their concentrations do not change.

For the reaction \[2 \mathrm{H}_2\mathrm{O}(g) \rightleftharpoons 2\mathrm{H}_2(g) + \mathrm{O}_2(g)\], the equilibrium constant expression is written as \[K = \frac{[\mathrm{H}_2(g)]^2 \times [\mathrm{O}_2(g)]}{([\mathrm{H}_2\mathrm{O}(g)])^2}\].

It is pivotal to raise the concentration of each species to the power of its coefficient in the balanced chemical equation. In this case, due to the stoichiometry, the concentration of hydrogen gas is squared, while the water vapor concentration is also squared in the denominator. Meticulously setting up this expression is the first step to solve related equilibrium problems effectively.

Stoichiometry deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. For gas phase reactions, stoichiometry becomes interesting as it involves volume, pressure, temperature, and the amount of gases (usually in moles).

Consider the reaction stoichiometry of \[2\mathrm{H}_2\mathrm{O}(g) \rightleftharpoons 2\mathrm{H}_2(g) + \mathrm{O}_2(g)\], where 2 moles of water vapor decompose to form 2 moles of hydrogen gas and 1 mole of oxygen gas. This relationship is key to determining the amount of products formed or reactants consumed when starting with a certain number of moles or volume of a gas at standard conditions.

When calculations are done based on the stoichiometry of the reaction, it’s crucial to ensure the balanced equation is used, as this represents the fixed ratios in which molecules react and form products. For students, visualizing molecules as individual units that react in these fixed ratios can simplify the concept and make the application of stoichiometry more intuitive.

In chemistry, molarity is a common unit for concentration, defined as the number of moles of solute per liter of solution (mol/L). Calculating molarity and moles is a foundational skill that enables us to understand the relationships between reactants and products in a chemical reaction.

For instance, once the equilibrium concentration of a gas is known, as in the formula \[[\mathrm{O}_2(g)] = 8.53 \times 10^{-3} M\], the next step is often to find out how many moles of this gas are present in a given volume of space. To find the moles, the molarity is multiplied by the volume in liters. Remember, the molarity formula is \[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{Volume of solution in liters}}\].

A practical tip for students is to always keep track of units throughout their calculations to avoid mistakes. For example, it’s essential to convert the volume into liters if it’s initially given in another unit. The clear understanding and correct application of molarity and moles calculation are vital for obtaining accurate results in chemical quantification.