Which starting condition would produce the larger volume of carbon dioxide by combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) with an excess of oxygen gas to produce carbon dioxide and water: (a) \(2.00 \mathrm{~L}\) of \(\mathrm{CH}_{4}(\mathrm{~g})\); (b) \(2.00 \mathrm{~g}\) of \(\mathrm{CH}_{4}(\mathrm{~g})\) ? Justify your answer. The system is maintained at a temperature of \(75^{\circ} \mathrm{C}\) and \(1.00\) atm.

The Ideal Gas Law is a crucial equation in chemistry and physics that allows us to calculate the relationship between the pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas. In mathematical terms, it is written as:
\( PV = nRT \).
This law implies that for a given amount of gas at constant temperature and pressure, the volume is directly proportional to the number of moles.
Let's apply this to our exercise. With the conditions given, the Ideal Gas Law helps us to determine how many moles of methane are in a 2.00 L volume at a specific temperature and pressure.
In our first step, after converting Celsius to Kelvin to use in the Ideal Gas Law, we can solve for the number of moles of methane. This step is critical because it lays the foundation for understanding how much reactant we have, which will then allow us to predict the volume of carbon dioxide produced from the combustion of methane.

Understanding molar mass is key to solving many chemistry problems, and it refers to the mass of one mole of a substance. It is expressed in grams per mole (g/mol) and is calculated by summing the atomic masses of all atoms in a molecule of the substance.
For methane (\( \text{CH}_4 \)), the molar mass is calculated by adding the mass of one carbon atom (12.01 g/mol) and the mass of four hydrogen atoms (4 x 1.008 g/mol). This gives us a molar mass of 16.04 g/mol for methane.
In our exercise, using the molar mass is essential in step 2, where we need to convert the mass of methane in grams to moles. This allows us to compare the amount of methane in moles whether we start with a mass in grams or a volume in liters, which is especially useful in stoichiometry.

Stoichiometry is the section of chemistry that pertains to quantifying the relationships between reactants and products in a chemical reaction. It relies on the balanced chemical equation and the mole concept to predict the amounts of substances consumed and produced in a reaction.
For the combustion of methane, the balanced equation is:
\( \text{CH}_4 + 2O_2 \rightarrow CO_2 + 2H_2O \).
This tells us that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water.
Using stoichiometry, we can relate the moles of methane to the moles of carbon dioxide produced in our problem. Since the equation shows a 1:1 molar ratio between methane and carbon dioxide, we know that the quantity of moles of methane we have (either from volume or mass) will be the same as the moles of carbon dioxide produced. This concept of proportional relationships is the heart of stoichiometry and is pivotal in solving for the larger volume of carbon dioxide produced in the exercise.

The chemical reaction yield is a measure of the efficiency of a reaction. It refers to the amount of product obtained from a chemical reaction compared to the amount that could be obtained if the reaction went to completion without any losses.
In a laboratory setting or industrial process, yields can be affected by factors such as incomplete reactions, side reactions, or loss of product during collection.
In our exercise, we are assuming that methane combusts completely with an excess of oxygen, implying a theoretical 100% yield. Thus, the amount of carbon dioxide produced can be directly predicted from the amount of methane consumed using stoichiometry.
The starting condition that produces more moles of methane will naturally lead to a higher theoretical yield of carbon dioxide. Consequently, in step 3 of our solution, by comparing the number of moles from both conditions—volume and mass—we can identify which yields a larger volume of carbon dioxide, assuming ideal behavior.