Problem 73 - Introductory Chemistry Atoms First

Question

Consider the balanced chemical equation \(2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (a) Given \(20.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) (hydrogen peroxide), how many grams of water will the reaction yield? (b) How many grams of hydrogen peroxide would you need to get \(20.0 \mathrm{~g}\) of water? (c) How many grams of hydrogen peroxide would you need to get \(20.0 \mathrm{~g}\) of \(\mathrm{O}_{2} ?\)

Step-by-Step Solution

Answer

(a) The reaction will yield \(10.6 \, g\) of water. (b) You would need \(37.7 \, g\) of hydrogen peroxide to get \(20 \, g\) of water. (c) You would need \(42.5 \, g\) of hydrogen peroxide to get \(20 \, g\) of \(\mathrm{O}_{2}\).

Step 1: Write down the balanced chemical equation

Given the balanced chemical equation: \[2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2}\]

Step 2: Calculate the molar masses of the substances involved

We will need to convert between moles and grams using the molar masses of the reactants and products. The molar mass of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is approximately \(34.0 \, g/mol\), of \(\mathrm{H}_{2} \mathrm{O}\) is approximately \(18.0 \, g/mol\), and of \(\mathrm{O}_{2}\) is approximately \(32.0 \, g/mol\). (a)

Step 3a: Convert given mass of hydrogen peroxide to moles

We are given \(20.0 \, g\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\). To convert to moles, we use the molar mass of hydrogen peroxide: \[\frac{20.0 \, g}{34.0 \, g/mol} = 0.588 \, mol\]

Step 4a: Use mole-to-mole ratio to find moles of water produced

From the balanced equation, we know that \(2 \, mol\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) yield \(2 \, mol\) of \(\mathrm{H}_{2} \mathrm{O}\). So, the moles of water produced are: \[0.588 \, mol \times \frac{2 \, mol \, \mathrm{H}_{2} \mathrm{O}}{2 \, mol \, \mathrm{H}_{2} \mathrm{O}_{2}} = 0.588 \, mol\]

Step 5a: Convert moles of water to grams

Finally, we will convert the moles of water produced to grams using the molar mass of water: \[0.588 \, mol \times 18.0 \, \frac{g}{mol} = 10.6 \, g\] So, the reaction will yield \(10.6 \, g\) of water. (b)

Step 3b: Convert given mass of water to moles

We are given \(20.0 \, g\) of \(\mathrm{H}_{2} \mathrm{O}\). To convert to moles, we use the molar mass of water: \[\frac{20.0 \, g}{18.0 \, g/mol} = 1.11 \, mol\]

Step 4b: Use mole-to-mole ratio to find moles of hydrogen peroxide needed

From the balanced equation, we know that \(2 \, mol\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced by \(2 \, mol\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\). So, the moles of hydrogen peroxide needed are: \[1.11 \, mol \times \frac{2 \, mol \, \mathrm{H}_{2} \mathrm{O}_{2}}{2 \, mol \, \mathrm{H}_{2} \mathrm{O}} = 1.11 \, mol\]

Step 5b: Convert moles of hydrogen peroxide to grams

Finally, we will convert the moles of hydrogen peroxide needed to grams using its molar mass: \[1.11 \, mol \times 34.0 \, \frac{g}{mol} = 37.7 \, g\] So, you would need \(37.7 \, g\) of hydrogen peroxide to get \(20 \, g\) of water. (c)

Step 3c: Convert given mass of oxygen to moles

We are given \(20.0 \, g\) of \(\mathrm{O}_{2}\). To convert to moles, we use the molar mass of oxygen: \[\frac{20.0 \, g}{32.0 \, g/mol} = 0.625 \, mol\]

Step 4c: Use mole-to-mole ratio to find moles of hydrogen peroxide needed

From the balanced equation, we know that one mole of \(\mathrm{O}_{2}\) is produced by \(2 \, mol\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\). So, the moles of hydrogen peroxide needed are: \[0.625 \, mol \times \frac{2 \, mol \,\mathrm{H}_{2} \mathrm{O}_{2}}{1 \, mol \, \mathrm{O}_{2}} = 1.25 \, mol\]

Step 5c: Convert moles of hydrogen peroxide to grams

Finally, we will convert the moles of hydrogen peroxide needed to grams using its molar mass: \[1.25 \, mol \times 34.0 \, \frac{g}{mol} = 42.5 \, g\] So, you would need \(42.5 \, g\) of hydrogen peroxide to get \(20 \, g\) of \(\mathrm{O}_{2}\).

Key Concepts

Molar Mass

Molar mass is a crucial concept in chemistry, particularly when dealing with stoichiometry. It refers to the mass of one mole of a substance, usually expressed in grams per mole (g/mol). For each compound, we calculate the molar mass by adding up the atomic masses of its constituent elements. This information is often found on the periodic table.
For example, to determine the molar mass of hydrogen peroxide (\(\mathrm{H}_{2} \mathrm{O}_{2}\)), we add the masses of 2 hydrogen atoms and 2 oxygen atoms:

• Hydrogen: 2 atoms at approximately 1.0 g/mol = 2.0 g/mol
• Oxygen: 2 atoms at approximately 16.0 g/mol = 32.0 g/mol
• Total molar mass of \(\mathrm{H}_{2} \mathrm{O}_{2}\) = 34.0 g/mol

This value is pivotal when converting between grams and moles, providing a bridge between the macroscopic and molecular worlds in chemical calculations.

Chemical Equation

Chemical equations are the language of chemistry, providing a shorthand way to describe what occurs in a chemical reaction. A balanced chemical equation shows the reactants converting into products in numeric ratios that maintain the conservation of mass.
Consider the equation for the decomposition of hydrogen peroxide:\[2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2}\]

• The numbers in front of the compounds, called coefficients, indicate the relative number of moles of each substance involved in the reaction.
• This equation tells us that 2 moles of \(\mathrm{H}_{2} \mathrm{O}_{2}\) decompose to form 2 moles of \(\mathrm{H}_{2} \mathrm{O}\) and 1 mole of \(\mathrm{O}_{2}\).
• Balancing equations ensures that the reaction obeys the law of conservation of mass, where the mass of reactants equals the mass of products.

Understanding this balance is essential for stoichiometric calculations, helping predict the amounts of products formed from given reactants.

Mole-to-Mole Ratio

The mole-to-mole ratio is a core principle of stoichiometry, derived directly from the coefficients of a balanced chemical equation. It tells us how many moles of one substance relate to another during a chemical reaction. This ratio is vital for converting between moles of reactants and products.
Let's look at the balanced equation of hydrogen peroxide decomposition again:\[2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2}\]

• The mole-to-mole ratio between \(\mathrm{H}_{2} \mathrm{O}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) is 1:1. For every 2 moles of hydrogen peroxide, 2 moles of water are produced.
• The ratio between \(\mathrm{H}_{2} \mathrm{O}_{2}\) and \(\mathrm{O}_{2}\) is 2:1. Thus, for every 2 moles of hydrogen peroxide, 1 mole of oxygen gas is produced.

This information allows chemists to predict the outcomes of reactions, ensuring the correct proportions of reactants to achieve desired products.

Hydrogen Peroxide Decomposition

Hydrogen peroxide decomposition is a classic example of a simple chemical reaction that illustrates important stoichiometric principles. It occurs when hydrogen peroxide (\(\mathrm{H}_{2} \mathrm{O}_{2}\)) breaks down into water (\(\mathrm{H}_{2} \mathrm{O}\)) and oxygen (\(\mathrm{O}_{2}\)). This decomposition is an exothermic reaction, meaning it releases energy in the form of heat.

• The balanced chemical equation for this reaction is: \(2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2}\)
• It serves as a practical demonstration of how chemical equations describe reactant-to-product transformations precisely.
• This process has natural and industrial relevance, occurring slowly at room temperature but can be sped up with a catalyst like manganese dioxide.

Understanding this decomposition teaches us about reactivity and the importance of catalysts in chemistry. Knowledge of such reactions is essential in fields ranging from medicine to rocket fuel technology.