The combustion of liquid ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) produces carbon dioxide and water. After \(3.8 \mathrm{~mL}\) of ethanol (density \(=0.789 \mathrm{~g} / \mathrm{mL}\) ) is allowed to burn in the presence of \(12.5 \mathrm{~g}\) of oxygen gas, \(3.10 \mathrm{~mL}\) of water (density \(=1.00 \mathrm{~g} / \mathrm{mL}\) ) is collected. Determine the limiting reactant, theoretical yield of \(\mathrm{H}_{2} \mathrm{O}\), and percent yield for the reaction. (Hint: Write a balanced equation for the combustion of ethanol.)

Understanding the concept of the limiting reactant is fundamental in chemistry for predicting the amount of product that can be formed in a chemical reaction. It refers to the reactant that will be used up first during the chemical process, thus determining when the reaction will stop. To identify the limiting reactant, one must compare the mole ratios of the reactants used in the reaction to those required by the balanced chemical equation.

In our exercise, we calculate that ethanol, \(C_2H_5OH\), is the limiting reactant because it will be consumed before the oxygen gas, \(O_2\). The reaction cannot proceed without the limiting reactant, even if other reactants are still available. By knowing which reactant limits the reaction, we can focus on this chemical to calculate theoretical and percent yields.

The theoretical yield is an essential aspect of stoichiometry that represents the maximum amount of product that can be generated as predicted by the balanced chemical equation, assuming complete reaction of the limiting reactant with no losses during the reaction. This ideal calculation is based entirely on stoichiometric ratios without factors like inefficiencies or side reactions. It is expressed in moles or grams and serves as a benchmark to calculate the efficiency of the actual experiment through percent yield.

In the given exercise, the theoretical yield of water, \(H_2O\), is calculated using the number of moles of the limiting reactant, ethanol, and the stoichiometric coefficients from the balanced equation. The result provides us with a target to measure how much water we theoretically expect to collect if everything were to react perfectly.

The percent yield is a measure of the efficiency of a chemical reaction. It compares the actual experimental yield to the theoretical yield and is expressed as a percentage. Calculating percent yield enables chemists to evaluate how well the reaction was carried out and to identify potential losses or side reactions. The formula for percent yield is:

\[\text{Percent yield} = \left(\frac{\text{Actual yield}}{\text{Theoretical yield}}\right) \times 100\%\].

In practice, the actual yield is often less than the theoretical yield due to factors like incomplete reactions, experimental errors, and product loss during extraction or purification. Our exercise demonstrates how to calculate the percent yield by dividing the mass of water collected after the reaction by the theoretically calculated mass of water.

Stoichiometry lies at the heart of chemical reactions. It is the part of chemistry that pertains to calculating the quantities of reactants and products involved in a chemical reaction. This includes determining the relative amounts needed to react without any excess and the expected quantities of outcomes from the process.

Stoichiometry relies on the law of conservation of mass and the balanced chemical equation to connect the mole ratios of reactants and products. In the provided solution, we use stoichiometry to transform the volume of ethanol into moles, calculate the required moles of oxygen, and further to turn the moles of ethanol into the moles, and then the mass, of water produced. This approach is the cornerstone for solving problems related to limiting reactants, theoretical yields, and percent yields.

A balanced chemical equation is critical for stoichiometric calculations as it provides the mole ratio of reactants to products. In a balanced equation, the number of atoms for each element is the same on both sides of the equation, respecting the law of conservation of mass. The coefficients in front of each compound indicate the ratio in which reactants combine to form products.

The exercise starts off by writing out the chemical equation for the combustion of ethanol: \[C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O\]. Here, the coefficients make it clear that one mole of ethanol reacts with three moles of oxygen to produce two moles of carbon dioxide and three moles of water. These ratios are then used throughout the solution to calculate the amounts of each substance involved and to understand the reaction's stoichiometry.