Cialculate the weight of lime (CaO) obtained by heating \(300 \mathrm{~kg}\) of \(90 \%\) pure limestone \(\left(\mathrm{CaCO}_{3}\right)\).

Understanding molar mass is essential to many stoichiometry problems. It represents the mass of one mole of a substance, usually in grams per mole (g/mol). The molar mass can be calculated by adding the atomic masses of the atoms in the molecular formula of the substance. To ensure clarity, let's look at how it was calculated for the substances in our exercise.

The molar mass of calcium carbonate \(\mathrm{CaCO}_3\) is the sum of the atomic mass of calcium (Ca), carbon (C), and three oxygen atoms (O). The atomic masses are found on the periodic table: for Ca it's approximately 40 g/mol, for C it's about 12 g/mol, and for O it's 16 g/mol. So, the molar mass of \(\mathrm{CaCO}_3\) is calculated as follows:\[\text{Molar mass of } \mathrm{CaCO}_3 = (40 + 12 + 3\times16) \text{ g/mol} = 100 \text{ g/mol}\].

This atomic mass reference needs to be accurate, thus taking a closer look at the periodic table, or using a molar mass calculator can ensure precision. Reiterating this step makes it clear to students that the seemingly simple task of adding atomic masses forms the backbone of many stoichiometry calculations.

Percentage purity is a key concept when you're dealing with chemical substances that are not 100% pure. It tells us how much of a mixture is the desired substance, which is particularly important when calculating how much of a product can be obtained from a reactant. In our exercise, the limestone is 90% pure, which means that only 90% of the total mass is actually calcium carbonate \(\mathrm{CaCO}_3\), and the rest is impurities.

To find the pure amount, we multiply the total mass by the percentage purity: \[\text{mass of pure }\mathrm{CaCO}_3 = 300\,\text{kg} \times 0.9 = 270\,\text{kg}\].

Purity can drastically affect the outcome of chemical reactions, so it's crucial to account for it when predicting yields. It's a simple multiplication, but its implications are vast, as ignoring purity can lead to significant discrepancies between theoretical and actual results.

Balancing chemical equations is a fundamental skill in chemistry. It's the process of ensuring that the number of each type of atom on the reactants side is the same as on the products side. This represents the conservation of mass and is crucial for accurately predicting the amounts of products formed in a chemical reaction.

In our exercise, the equation was straightforward to balance because the decomposition of calcium carbonate \(\mathrm{CaCO}_3\) into calcium oxide (CaO) and carbon dioxide \(\mathrm{CO}_2\) naturally balances itself:\[\mathrm{CaCO}_3 \rightarrow \mathrm{CaO} + \mathrm{CO}_2\].

Each side of the equation has one calcium atom, one carbon atom, and three oxygen atoms. It's crucial to check chemical equations for balance before using them in stoichiometry calculations; an unbalanced equation can lead to incorrect outcomes.

The theoretical yield is the amount of product that can be made in a chemical reaction based on the amount of limiting reactant. It's a calculation based on stoichiometry – the quantitative relationship between reactants and products in a chemical reaction.

In our exercise, once we found the pure mass of \(\mathrm{CaCO}_3\) and the molar masses of \(\mathrm{CaCO}_3\) and CaO, we used stoichiometry to find the theoretical mass of CaO produced:\[\text{mass of CaO} = (\text{mass of pure }\mathrm{CaCO}_3 / \text{molar mass of }\mathrm{CaCO}_3) \times \text{molar mass of CaO}\].

This involved using the ratio of moles from the balanced equation, which in this case is a 1:1 ratio, leading to a straightforward calculation. Understanding how to determine the theoretical yield is crucial for planning and analyzing chemical reactions. It serves as a benchmark for comparing against the actual yield obtained from a chemical process.