Hydrogen peroxide with a concentration of 3.0 percent \(\left(3.0 \mathrm{~g}\right.\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) in \(100 \mathrm{~mL}\) of solution) is sold in drugstores for use as an antiseptic. For a \(10.0-\mathrm{mL}\) 3.0 percent \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution, calculate (a) the oxygen gas produced (in liters) at STP when the compound undergoes complete decomposition and (b) the ratio of the volume of \(\mathrm{O}_{2}\) collected to the initial volume of the \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution.

Understanding the molar mass of a compound is crucial for diving into any kind of quantitative chemical problem-solving. Molar mass refers to the weight of one mole (6.022 \( \times \) 10²³ entities) of a substance and is expressed in grams per mole (g/mol).

To calculate the molar mass, one simply needs to sum the atomic masses of all the atoms in a molecule. For instance, hydrogen peroxide (H₂O₂) is composed of two hydrogen atoms and two oxygen atoms. Taking into account that the atomic mass of hydrogen is approximately 1 g/mol and of oxygen is about 16 g/mol, the molar mass of H₂O₂ comes out to (2 \( \times \) 1) + (2 \( \times \) 16) = 34 g/mol.

This figure is fundamental in the step by step solution provided, allowing us to convert the mass of hydrogen peroxide into moles, which is the gateway to further stoichiometric calculations.

Stoichiometry is the section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.

To operate with stoichiometry, it's essential to start with a balanced chemical equation. This tells us the proportion in which reactants combine and products form. Taking the decomposition of hydrogen peroxide as an example, the balanced equation is:2 H₂O₂ → 2 H₂O + O₂. It indicates that two molecules of hydrogen peroxide yield two molecules of water and one molecule of oxygen gas.

The stoichiometric coefficients (the numbers in front of molecules) inform us of the ratio of reactants to products. In the given exercise, having 0.0088 moles of H₂O₂, we can use these ratios to calculate the moles of oxygen gas produced.

Calculating the volume of a gas at standard temperature and pressure (STP) is simplified by the fact that any gas at STP occupies 22.4 liters per mole. STP is defined as a temperature of 273.15 Kelvin (0°C) and a pressure of 1 atmosphere.

In practical terms, if a chemical reaction at STP produces a gas, we can directly relate the quantity in moles to its volume. For oxygen gas (O₂), which has a molar volume of 22.4 L/mol at STP, you just multiply the amount of moles by 22.4 L/mol to obtain the volume it occupies.

Real-life Application

For instance, your solved exercise detailed that 0.0044 moles of O₂ gas were produced. Applying the molar volume, we discover that this amount corresponds to 0.0044 moles \( \times \) 22.4 L/mol = 0.099 liters of oxygen at STP. This is a straightforward way to transition from abstract moles to a tangible volume that can be measured and applied.

Balancing chemical equations is akin to an art form in chemistry, ensuring that the same number of each type of atom exists on both sides of the reaction. This is a manifestation of the law of conservation of mass. To balance an equation, we adjust the coefficients in front of the chemical formulas until the number of atoms of each element is equal on both sides.

When we say an equation is balanced, it means that it reflects the stoichiometric proportions of a chemical reaction accurately. For example, the decomposition of hydrogen peroxide is shown by the balanced chemical equation: 2 H₂O₂ → 2 H₂O + O₂. It implies two moles of hydrogen peroxide yield one mole of oxygen gas, a fact that was utilized in your step by step solution.

Getting the balance right is the foundation for any stoichiometric calculation. Without a properly balanced equation, the ensuing calculations may not reflect the real-world quantities and ratios involved in the reaction.