Consider the reaction: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) $$ A reaction flask initially contains \(0.10\) atm of \(\mathrm{SO}_{2}\) and \(0.10\) atm of \(\mathrm{O}_{2}\). What is the total pressure in the flask once the limiting reactant is completely consumed? Assume a constant temperature and volume and a \(100 \%\) reaction yield.

Understanding the concept of a limiting reactant is crucial in the study of chemical reactions. In simple terms, the limiting reactant is the substance in a chemical reaction that is entirely consumed first, stopping the reaction from continuing because there is no more of it left to react. It's akin to running out of flour when baking a cake; no matter how much of the other ingredients you have, without enough flour, you can't make more cake.

In the given exercise involving the reaction between sulfur dioxide (SO2) and oxygen (O2) to form sulfur trioxide (SO3), the limiting reactant is essential for calculating the total pressure inside the reaction flask after the reaction has occurred. If we mistake which reactant is limiting, our calculations for the final pressure would be incorrect. The key is to compare the mole ratio of the reactants to the ratio required by the balanced chemical equation to identify which reactant will run out first.

Stoichiometry allows us to predict the outcomes of chemical reactions in terms of the amounts of reactants and products involved. It is the methodology used to determine the proportions in which chemicals react according to the balanced chemical equation. It's much like a recipe that tells you how many cups of each ingredient you need to bake a cake.

For the reaction at hand, stoichiometric calculations tell us that 2 moles of SO2 reacts with 1 mole of O2 to produce 2 moles of SO3. This ratio is the foundation for determining how much SO3 will be produced from the initial pressure of SO2. In our exercise, the assumption of a 100% reaction yield simplifies the calculation, allowing us to directly equate the pressure of SO2 used to the pressure of SO3 produced. When approaching stoichiometric problems, always ensure the chemical equation is balanced first and then apply the mole ratios accordingly.

The partial pressure of a gas is the pressure it would exert if it alone occupied the entire volume of the mixture at the same temperature. Imagine you have several different gases inside a balloon; each one pushes against the balloon's rubber with a certain amount of force. This force for an individual type of gas is its partial pressure, and the sum of all these forces gives the total pressure inside the balloon.

In our exercise, the initial pressures of SO2 and O2 contribute to the total pressure inside the flask. As the reaction proceeds to completion, the partial pressure of the limiting reactant, SO2, drops to zero, since it is entirely consumed. The partial pressure of the product, SO3, now needs to be considered, along with any remaining partial pressure from the excess reactant, O2. By adding the partial pressure of SO3 and the leftover partial pressure of O2, we arrive at the total pressure in the flask after the completion of the reaction.