Consider the following unbalanced chemical equation for the combustion of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right):\) $$\mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ If \(20.4 \mathrm{g}\) of pentane are burned in excess oxygen, what mass of water can be produced, assuming \(100 \%\) yield?

Understanding the process of balancing chemical equations is foundational to solving many problems in chemistry, including those involving stoichiometry.
Balancing chemical equations is akin to ensuring that the same number of atoms for each element is present on both sides of the reaction. It's like a mathematical equation where the principle of conservation of mass must be respected.
When we look at our exercise example of pentane combustion, we initially have an unbalanced equation with the formula \(\mathrm{C}_{5} \mathrm{H}_{12}(l) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l)\). This step is crucial, as it sets the stage for accurately determining the amounts of reactants and products involved.

Another critical step in solving stoichiometry problems is calculating the molar mass of substances. The molar mass is the weight of one mole of a substance, typically expressed in grams per mole (g/mol).
It is vital for converting between mass and moles of a substance. In the case of our combustion exercise, the molar mass of pentane (\(\mathrm{C}_{5}\mathrm{H}_{12}\)) is calculated by multiplying the atomic masses of carbon (\(12.01\ \text{g/mol}\)) and hydrogen (\(1.01\ \text{g/mol}\)) by the number of atoms in one molecule and summing them up. This step allows us to convert the given mass of pentane into moles, which is a necessary conversion to use in stoichiometry calculations.

The yield of a chemical reaction is a measure of how efficient the reaction is at producing the desired product. It's often expressed as a percentage and is calculated by comparing the amount of product actually obtained to the amount that could have been produced in a perfect, 100% yield reaction.
In real laboratory conditions, the yield is rarely 100% due to various factors such as incomplete reactions, side reactions, or loss of product during the process. However, for the purpose of our stoichiometry exercise, it's assumed that the reaction has a 100% yield. This simplification is common in textbook problems to focus on the stoichiometry without the added complexity of real-world inefficiencies. By understanding these three core concepts, students become better equipped to tackle stoichiometric calculations and comprehend the quantitative aspects of chemical reactions.