When \(0.031\) of a mixture of hydrogen and oxygen was exploded, \(0.0031\) of oxygen remained. The initial mixture contains (by volume) (a) \(60 \% \mathrm{O}_{2}\) (b) \(40 \% \mathrm{O}_{2}\) (c) \(50 \% \mathrm{O}_{2}\) (d) \(30 \% \mathrm{O}_{2}\)

Chemical reaction equations are the foundation for understanding stoichiometry problems. They provide critical information about the reactants and products involved in a chemical reaction, as well as their proportions. For example, the balanced reaction equation for the combustion of hydrogen gas is \(2H_2 + O_2 \rightarrow 2H_2O\).

This equation indicates that two molecules of hydrogen gas (\(H_2\)) react with one molecule of oxygen gas (\(O_2\)) to form two molecules of water (\(H_2O\)). Notice the coefficients in front of the chemical formulas which indicate the molar ratio in which substances react. In this case, the reaction requires a 2:1 ratio of hydrogen to oxygen. This molar ratio is critical when performing stoichiometry calculations, particularly if you are given a mass of one reactant and need to determine the mass of another.

To solve stoichiometry problems, understanding molar mass is crucial. The molar mass of a substance is the mass of one mole of that substance, commonly expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in one molecule of the substance.

For instance, the molar mass of water (\(H_2O\)) is 18 g/mol because each molecule has two hydrogen atoms (with a molar mass of approximately 1 g/mol each) and one oxygen atom (with a molar mass of approximately 16 g/mol). When working with chemical reaction equations, you'll often calculate the amount of a reactant or product using its molar mass. For example, if you are given the mass of hydrogen that reacted, you can find the moles of hydrogen by dividing the mass by its molar mass (2 g/mol), as shown in the exercise's solution steps.

Percentage composition by volume is another important concept in stoichiometry. It tells us the volume of each component as a percentage of the total mixture's volume. Especially in gas mixtures, this can be a useful indication of the composition because gases tend to occupy the same volume under the same conditions of temperature and pressure, regardless of the species.

In the context of the given problem, after finding the initial masses and the molar masses of hydrogen and oxygen, we calculated their moles. The next crucial step was to determine the total moles in the mixture, which allowed us to calculate each gas's percentage by volume. It's important to remember that the percentage of a component is computed by dividing the moles of the component by the total moles of all components and then multiplying by 100%. This process is clearly illustrated in the solution's steps for both hydrogen and oxygen, leading to the percentage composition of each gas within the original mixture.