Question

How many grams of hydrogen peroxide must decompose to give \(55.0 \mathrm{~mL}\) of oxygen gas at STP? $$\mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)$$

Short answer

0.167 grams of hydrogen peroxide must decompose.

Step-by-step solution

Understanding STP conditions

STP (Standard Temperature and Pressure) is defined as a temperature of 273.15 Kelvin (0°C) and a pressure of 1 atm. Under these conditions, 1 mole of any ideal gas occupies a volume of 22.4 liters (or 22,400 mL).

Calculate moles of oxygen gas

To find the moles of oxygen gas, we use the relation that 1 mole of a gas at STP occupies 22,400 mL. Therefore, to find the moles of oxygen in 55.0 mL, use the formula: \[ ext{Moles of } O_2 = rac{55.0 ext{ mL}}{22,400 ext{ mL/mol}} \] Evaluating this gives: \[ ext{Moles of } O_2 = 0.002455 ext{ mol} \]

Stoichiometry of the reaction

From the balanced chemical equation: \[ ext{2 } ext{H}_2 ext{O}_2 (l) ightarrow ext{2 } ext{H}_2 ext{O} (l) + ext{O}_2 (g) \] We see that 2 moles of \(\text{H}_2\text{O}_2\) decompose to produce 1 mole of \(\text{O}_2\). Thus, the moles of \(\text{H}_2\text{O}_2\) needed for 0.002455 moles of \(\text{O}_2\) is: \[ 2 imes 0.002455 = 0.00491 ext{ moles of } \text{H}_2\text{O}_2 \]

Calculate mass of hydrogen peroxide

The molar mass of \(\text{H}_2\text{O}_2\) is approximately 34.02 g/mol (1.01 * 2 + 16.00 * 2). Hence, the gram amount is calculated by: \[ ext{Mass of } \text{H}_2\text{O}_2 = 0.00491 ext{ mol} imes 34.02 ext{ g/mol} \] Evaluating this gives: \[ ext{Mass of } \text{H}_2\text{O}_2 = 0.167 ext{ g} \]

Key concepts

Mole Concept

The mole concept is a fundamental idea in chemistry that allows us to count particles by weighing them. One mole of any substance contains exactly Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) particles. This allows chemists to convey and convert between atoms, ions, molecules, or moles efficiently. When dealing with gases, like in our exercise, it becomes particularly useful because one mole of a gas at Standard Temperature and Pressure (STP) occupies 22.4 liters (22,400 mL). This relationship lets us calculate the number of moles from a given volume, making it easier to work with reactions involving gases at STP. To find the moles of a gas at STP, use the formula:

• Moles \( = \frac{\text{Volume in mL}}{\text{22,400 mL/mol}}\)

In the exercise, finding moles of \(\text{O}_2\) allows us to backtrack and find how much \(\text{H}_2\text{O}_2\) is needed to produce it.

Ideal Gas Law

The Ideal Gas Law is an equation of state that describes how gases behave under various conditions. It combines several gas laws, including Boyle's, Charles's, and Avogadro's laws. The equation is expressed as:\[ PV = nRT \]Where:

• \(P\) is the pressure in atm
• \(V\) is the volume in liters
• \(n\) is the number of moles
• \(R\) is the ideal gas constant \(0.0821 \frac{L \cdot atm}{K \cdot mol}\)
• \(T\) is the temperature in Kelvin

In our exercise, the conditions are at STP, which is useful because it simplifies calculations by allowing us to directly use the 22.4 L/mol relationship for gases. Understanding this helps when calculating the moles from given volumes of a gas at STP, as demonstrated in calculating the moles of oxygen from its volume.

Chemical Reactions

Chemical reactions involve rearranging atoms to convert reactants into products. In the exercise, the decomposition of hydrogen peroxide \(\text{H}_2\text{O}_2\) into water \(\text{H}_2\text{O}\) and oxygen \(\text{O}_2\) is the process we are analyzing. The balanced chemical equation is vital as it shows the stoichiometry of the reaction:\[ 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \]From this equation, we see that we need 2 moles of hydrogen peroxide to produce 1 mole of oxygen gas. This ratio helps in determining how much reactant is necessary to produce a given quantity of product. In the exercise, using this stoichiometry, we calculate how many moles and, subsequently, the mass of \(\text{H}_2\text{O}_2\) required to result in the specified amount of \(\text{O}_2\). By understanding the molar ratios from balanced reactions, we can solve many real-world chemical problems.