Question

Hydrogen peroxide is used as a cleansing agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes on contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams on contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide: $$\mathrm{BaO}_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{BaCl}_{2}(a q)$$ What mass of hydrogen peroxide should result when \(1.50 \mathrm{g}\) barium peroxide is treated with \(25.0 \mathrm{mL}\) hydrochloric acid solution containing \(0.0272 \mathrm{g}\) HCl per mL? What mass of which reagent is left unreacted?

Short answer

In conclusion, when 1.50 g of barium peroxide is treated with 25.0 mL of hydrochloric acid solution containing 0.0272 g HCl per mL, 0.301 g of hydrogen peroxide is formed and 0.0328 g of hydrochloric acid remains unreacted.

Step-by-step solution

Calculate the moles of reactants

We need to find the moles of the given reactants, barium peroxide (BaO2) and hydrochloric acid (HCl). First, calculate the molar mass of barium peroxide: BaO2: \(137.33 \ g/mol (Ba) + 2 * 16.00 \ g/mol (O) = 169.33 \ g/mol\) Now, calculate the moles of barium peroxide: Moles of BaO2 = \(\frac{1.50 \ g}{169.33 \ g/mol} = 0.00886 \ mol\) Next, calculate the moles of hydrochloric acid using the given concentration and volume: Moles of HCl = \(25.0 \ mL * \frac{0.0272 \ g}{1 \ mL} * \frac{1 \ mol}{36.46 \ g/mol} = 0.0186 \ mol\)

Determine the limiting reactant

To find the limiting reactant, we must compare the moles of the reactants with the stoichiometry of the balanced chemical equation. From the balanced equation, we know that 1 mole of BaO2 reacts with 2 moles of HCl. So, we must determine if we have a sufficient amount of HCl for the given amount of BaO2. Using the stoichiometric ratio: Moles of HCl required for BaO2 reaction: \(0.00886 \ mol (BaO2) * \frac{2 \ mol (HCl)}{1 \ mol (BaO2)} = 0.0177 \ mol\) Since we have 0.0186 moles of HCl, there is enough HCl to react with all of the given BaO2. Thus, BaO2 is the limiting reactant, while HCl is the excess reactant and some amount will be left unreacted.

Calculate the mass of hydrogen peroxide formed

To calculate the mass of hydrogen peroxide (H2O2) formed, first determine the moles of H2O2 generated using the stoichiometry of the balanced equation. Moles of H2O2 = Moles of BaO2 (limiting reactant) = 0.00886 mol Next, calculate the molar mass of hydrogen peroxide: H2O2: \(2 * 1.01 \ g/mol (H) + 2 * 16.00 \ g/mol (O) = 34.02 \ g/mol\) Now, find the mass of hydrogen peroxide produced: Mass of H2O2 = \(0.00886 \ mol * 34.02 \ g/mol = 0.301 \ g\)

Calculate the mass of the excess reactant left unreacted

To find the mass of the excess reactant (HCl) left unreacted, first determine the moles of HCl that reacted with the limiting reactant (BaO2). Moles of HCl reacted = \(0.00886 \ mol (BaO2) * \frac{2 \ mol (HCl)}{1 \ mol (BaO2)} = 0.0177 \ mol\) Subtract the moles of HCl that reacted from the initial moles of HCl to find the moles of HCl left unreacted: Moles of HCl unreacted = \(0.0186 \ mol - 0.0177 \ mol = 0.0009 \ mol\) Now, calculate the mass of the unreacted HCl: Mass of HCl unreacted = \(0.0009 \ mol * 36.46 \ g/mol = 0.0328 \ g\) In conclusion, the mass of hydrogen peroxide formed is 0.301 g, and the mass of unreacted hydrochloric acid is 0.0328 g.

Key concepts

Stoichiometry Fundamentals

Stoichiometry is a key concept in chemistry that involves calculating the relationships between the amounts of reactants and products in a chemical reaction. In the given reaction:

• Barium peroxide (\(\text{BaO}_2\)) reacts with hydrochloric acid (\(\text{HCl}\)) to form hydrogen peroxide (\(\text{H}_2\text{O}_2\)) and barium chloride (\(\text{BaCl}_2\)).

To understand stoichiometry, we need to know the balanced chemical equation. This provides the ratio of reactants to products, allowing us to calculate necessary quantities.
The balanced equation:
\[\text{BaO}_{2}(s)+2 \text{HCl}(aq) \rightarrow \text{H}_2\text{O}_{2}(aq)+\text{BaCl}_2(aq)\]
This means 1 mole of \(\text{BaO}_2\) reacts with 2 moles of \(\text{HCl}\), helping us compute how much \(\text{H}_2\text{O}_2\) can be produced. Stoichiometry is like a recipe, telling us how to mix our ingredients for just the right result.

Understanding Limiting Reactant

In every chemical reaction, a limiting reactant is the substance that is completely consumed first, limiting the amount of product formed. To find it:

• First, calculate moles of each reactant (\(\text{BaO}_2\) and \(\text{HCl}\)).

When comparing these moles to the reaction's stoichiometric ratios, the reactant in lesser proportion becomes the limiting one.
In this exercise, \(\text{BaO}_2\) was the limiting reactant because it produced less product, thereby fully reacting before \(\text{HCl}\) was used up. Once we know the limiting reactant, we can calculate the maximum possible yield of the product, in this case, hydrogen peroxide.

Grasping Molar Mass

Molar mass is the weight of one mole of a substance, fundamental for converting between mass and moles in chemistry. It's often expressed as grams per mole (g/mol).

• For \(\text{BaO}_2\): Calculating as \(137.33 \ g/mol (\text{Ba}) + 2 \times 16.00 \ g/mol (\text{O}) = 169.33 \ g/mol\).

Knowing the molar mass allows us to calculate the number of moles from the given mass. This is crucial for determining how much of each reactant is present, and ultimately, how much product can form.
Using molar mass simplifies solving stoichiometry problems by connecting measurable quantities to abstract mole calculations.

Role of Oxidizing Agents

Oxidizing agents are substances that can accept electrons in a chemical reaction, often leading to the oxidation of another substance.

• In the body of reactions like ours, hydrogen peroxide (\(\text{H}_2\text{O}_2\)) serves as an effective oxidizing agent.

It can kill microorganisms, making it useful in medical and laboratory settings for its antibacterial properties.
This characteristic stems from its ability to release oxygen upon decomposition. When involved in reactions, oxidizing agents play a crucial role in driving the process by interacting with other substances, facilitating the breakdown or transformation of materials.