What volume of oxygen, measured at \(27^{\circ} \mathrm{C}\) and 101.325 \(\mathrm{kPa}\) , is needed for the combus- tion of 1.11 \(\mathrm{kg}\) of coal? (Assume coal is 100\(\%\) carbon.) $$\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g$$

Stoichiometry is a section of chemistry that involves using relationships within a chemical equation to deduce the quantitative data of reactants and products. In the context of gases, the Ideal Gas Law is paramount and can be expressed with the equation \( PV = nRT \). This law allows us to calculate the volume \( (V) \) of a gas when temperature \( (T) \) , pressure \( (P) \) , and the number of moles \( (n) \) are known, with \( R \) being the gas constant.

For example, to determine the volume of oxygen gas needed for the combustion of coal, we apply this law. We begin by converting the mass of a substance — in this case, coal, which is assumed to be pure carbon — to moles. Here, the provided 1.11 kg of coal is first converted to grams and then to moles using carbon's molar mass. Following this, we use stoichiometry to equate the moles of carbon with the moles of oxygen needed, adhering to the chemical reaction's 1:1 molar ratio. Lastly, by entering these values into the Ideal Gas Law formula, along with the temperature (converted to Kelvin) and the given pressure, we obtain the volume of oxygen gas required.

The term molar mass is used to express the mass of one mole of a substance, usually in grams per mole (g/mol). Calculating molar mass is crucial in converting between grams of a substance and moles, which is a common step in stoichiometry.

For instance, in our exercise, we convert the mass of coal, which is pure carbon in this scenario, to moles using the molar mass of carbon. The calculation goes as follows: \[ \text{Moles of C} = \frac{\text{Mass of C in grams}}{\text{Molar Mass of C}} \. \] By using the molar mass of carbon (12.01 g/mol), we accurately determine how many moles of carbon are present in the given weight, which then feeds into further stoichiometric conversions.

The balancing of a chemical reaction is essential for correct stoichiometric calculations. A balanced chemical equation ensures that the law of conservation of mass is adhered to, meaning that the number of atoms of each element on the reactant side equals the number of atoms of the same element on the product side.

In the combustion of carbon to produce carbon dioxide, we balance the reaction by ensuring that there are equal numbers of each type of atom before and after the reaction: \( \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) \). Here, one atom of carbon reacts with one molecule of oxygen to produce one molecule of carbon dioxide, thus a 1:1:1 ratio. This balance allows us to deduce that for each mole of carbon combusted, one mole of oxygen is required. As such, for stoichiometric calculations, it is imperative to start with a balanced equation to get accurate results.