Question

Given the decomposition of hydrogen peroxide, calculate the moles of oxygen gas produced from \(5.00 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\). $$2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)$$

Short answer

2.50 moles of \( \mathrm{O_2} \) are produced from 5.00 moles of \( \mathrm{H_2O_2} \).

Step-by-step solution

Understanding the Reaction

The decomposition of hydrogen peroxide, \( \mathrm{H_2O_2} \), yields water \( \mathrm{H_2O} \) and oxygen gas \( \mathrm{O_2} \). According to the balanced chemical equation: \( 2\mathrm{H_2O_2} \rightarrow 2\mathrm{H_2O} + \mathrm{O_2} \), two moles of hydrogen peroxide decompose to produce one mole of oxygen gas.

Determine Mole Ratio

From the balanced equation, note the molar ratio between \( \mathrm{H_2O_2} \) and \( \mathrm{O_2} \). For every 2 moles of \( \mathrm{H_2O_2} \) decomposed, 1 mole of \( \mathrm{O_2} \) is produced.

Calculate Moles of Oxygen Gas

Using the molar ratio from the balanced chemical equation, calculate the moles of \( \mathrm{O_2} \) produced from 5.00 moles of \( \mathrm{H_2O_2} \). Use the formula: \( \frac{5.00 \, \text{mol of } \mathrm{H_2O_2}}{2 \text{ moles of } \mathrm{H_2O_2}} = x \,\text{mol of } \mathrm{O_2} \), which simplifies to \( x = 2.50 \, \text{mol of } \mathrm{O_2} \).

Key concepts

Chemical Reactions

Chemical reactions are processes where reactants are transformed into products. In the case of hydrogen peroxide (H_2O_2), it decomposes into water (H_2O) and oxygen gas (O_2). These transformations involve breaking bonds in the reactants and forming new bonds to produce the products.
Reactants and products are represented in chemical equations, which provide a concise way to describe these molecular changes. These equations show both the formulas of the involved substances and the proportion in which they react or form.
In our given reaction: \[2 \, ext{H}_2 ext{O}_2 ightarrow 2 \, ext{H}_2 ext{O} + ext{O}_2 \] two moles of hydrogen peroxide decompose into two moles of water and one mole of oxygen gas. This kind of equation helps in identifying how much of each substance is used and produced in the reaction.

Molecular Calculations

Molecular calculations are crucial for understanding how much of a substance is involved in a chemical reaction. Key to this is using quantities like moles, which are a standard way to measure amount of substance in chemistry.
Moles relate directly to the matter's molecular structure through Avogadro's number, which is approximately \( 6.022 \times 10^{23} \). This allows chemists to count specific entities, such as molecules or atoms, in a given mass of substance.
In our example, we calculate moles to find out how much oxygen gas (O_2) is formed from the decomposition of a given amount of hydrogen peroxide (H_2O_2). According to the balanced chemical equation:

• For every 2 moles of (H_2O_2), 1 mole of (O_2) is produced.

Given 5.00 moles of (H_2O_2), we use \[\frac{5.00 \, ext{mol} \, ext{H}_2 ext{O}_2}{2 \, ext{mol} \, ext{H}_2 ext{O}_2/ ext{mol} \, ext{O}_2} = 2.50 \, ext{mol} \, ext{O}_2 \]This means 2.50 moles of oxygen gas is produced, clearly showing the conversion from reactant to product.

Balanced Chemical Equations

Balanced chemical equations are a fundamental concept in stoichiometry. They ensure the law of conservation of mass is followed in chemical reactions, meaning the same number of each type of atom is present on both sides of the equation.
For the decomposition of hydrogen peroxide:\[2 \, ext{H}_2 ext{O}_2 ightarrow 2 \, ext{H}_2 ext{O} + ext{O}_2\] The equation is balanced because it has:

• Four hydrogen atoms on each side (2 moles of H_2O_2 = 4 hydrogens, 2 moles of H_2O = 4 hydrogens)
• Four oxygen atoms on each side (2 moles of H_2O_2 = 4 oxygens, 2 moles of H_2O + 1 mole O_2 = 4 oxygens)

Balancing equations is crucial as it helps determine the correct proportions of reactants and products. It also assists in accurately forecasting the quantities of substances required or standardized product amounts, which is essential for scaling reactions in experimental and industrial settings.