Outline the steps needed to determine the limiting reactant when \(30.0 \mathrm{g}\) of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\), is bumed with \(75.0 \mathrm{g}\) of oxygen. Determine the limiting reactant.

Stoichiometry is a section of chemistry that involves calculations of the reactants and products in a chemical reaction. It helps predict the amounts of substances consumed and produced in a given reaction. The central principle of stoichiometry is the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. Consequently, the same number of atoms must be present before and after the reaction.

To perform stoichiometric calculations, you first need the balanced chemical equation, which provides the mole ratio of reactants and products. This mole ratio is essential for calculating how much of each reactant is needed to produce a certain amount of product and to determine the limiting reactant—the substance that will be used up first, restricting the amount of product that can form.

For improved understanding, it's helpful to visualize stoichiometry like a recipe; just as a recipe specifies the amount of each ingredient needed to make a certain number of servings, a balanced chemical equation indicates how much of each reactant is needed to produce a set amount of product.

Balancing chemical equations is a fundamental aspect of stoichiometry and ensures adherence to the law of conservation of mass. To balance a chemical equation, each type of atom on the reactant side must be equal to the number of atoms of the same type on the product side.

The balanced equation in our exercise, \( \mathrm{C}_{3} \mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O} \), visually represents a scale in equilibrium. To achieve this balance, coefficients are strategically placed before chemical formulas to indicate the number of molecules (or moles) involved in the reaction.

When balancing equations, it is usually easiest to start with the most complex molecule, which in this case is propane (\(\mathrm{C}_{3} \mathrm{H}_{8}\)). After balancing carbon and hydrogen atoms, the last step is to adjust the oxygen atoms, often found in the diatomic oxygen molecule (\(\mathrm{O}_{2}\)). \(H_{4}\) headlines can be useful for breaking down the step-by-step process of balancing complex chemical equations, ensuring a comprehensive understanding.

Molar mass is a physical property expressed as the mass of a given substance (chemical element or chemical compound) divided by its amount of substance. The unit for molar mass is grams per mole (\( \text{g/mol} \)).

To calculate the molar mass, you look at the formula of the compound and sum up the atomic weights of all the atoms in the formula. The atomic weights are found on the periodic table and are typically reported in atomic mass units (amu), which correlates directly to grams per mole for these calculations. It is crucial to account for the number of atoms of each element in the compound and multiply the atomic weight by this number.

In the exercise, the calculation for propane involves adding thrice the atomic weight of carbon to eight times the atomic weight of hydrogen. Understanding molar mass is vital for converting grams to moles, which is necessary when using the balanced chemical equation to find out how much of each substance is involved in the reaction.