Problem 16 - Chemistry Matter and Change

Question

Challenge In the formation of acid rain, sulfur dioxide \(\left(\mathrm{SO}_{2}\right)\) reacts with oxygen and water in the air to form sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right) .\) Write the balanced chemical equation for the reaction. If 2.50 \(\mathrm{g}\) of \(\mathrm{SO}_{2}\) reacts with excess oxygen and water, how much \(\mathrm{H}_{2} \mathrm{SO}_{4}\) , in grams, is produced?

Step-by-Step Solution

Answer

3.82 g of \(\mathrm{H}_{2}SO_{4}\) is produced.

Step 1: Write Unbalanced Equation

Identify the reactants and products in the reaction. The unbalanced equation is: \[ \mathrm{SO}_{2} + \mathrm{O}_{2} + \mathrm{H}_{2}O \rightarrow \mathrm{H}_{2}SO_{4} \]

Step 2: Balance Chemical Equation

Balance the reaction by adjusting coefficients. The balanced equation becomes: \[ 2\,\mathrm{SO}_{2} + \mathrm{O}_{2} + 2\,\mathrm{H}_{2}O \rightarrow 2\,\mathrm{H}_{2}SO_{4} \] This ensures mass conservation: 2 sulfur, 6 oxygen, and 4 hydrogen atoms on each side.

Step 3: Calculate Molar Mass of SO2

Find the molar mass of \(\mathrm{SO}_{2}\). Sulfur has an atomic mass of 32.07 g/mol and oxygen has 16.00 g/mol. So, \(\text{Molar mass of } \mathrm{SO}_{2} = 32.07 + 2 \times 16.00 = 64.07\, \text{g/mol}\).

Step 4: Convert SO2 Mass to Moles

Convert the mass of \(\mathrm{SO}_{2}\) to moles using its molar mass. \[ \text{Moles of } \mathrm{SO}_{2} = \frac{2.50\, \text{g}}{64.07\, \text{g/mol}} = 0.039\, \text{mol} \]

Step 5: Use Stoichiometry to Find Moles of H2SO4

Use the balanced equation to find moles of \(\mathrm{H}_{2}SO_{4}\) produced from \(\mathrm{SO}_{2}\). The equation ratio is 2 moles of \(\mathrm{SO}_{2}\) to 2 moles of \(\mathrm{H}_{2}SO_{4}\). Thus, \(0.039\, \text{mol}\) of \(\mathrm{SO}_{2}\) produces \(0.039\, \text{mol}\) of \(\mathrm{H}_{2}SO_{4}\).

Step 6: Calculate Molar Mass of H2SO4

Determine the molar mass of \(\mathrm{H}_{2}SO_{4}\). It consists of 2 hydrogen atoms (1.01 g/mol each), 1 sulfur atom (32.07 g/mol), and 4 oxygen atoms (16.00 g/mol each). \[ \text{Molar mass of } \mathrm{H}_{2}SO_{4} = 2 \times 1.01 + 32.07 + 4 \times 16.00 = 98.09\, \text{g/mol} \]

Step 7: Convert Moles of H2SO4 to Mass

Convert moles of \(\mathrm{H}_{2}SO_{4}\) to grams using its molar mass. \[ \text{Mass of } \mathrm{H}_{2}SO_{4} = 0.039\, \text{mol} \times 98.09\, \text{g/mol} = 3.82\, \text{g} \]

Key Concepts

Stoichiometry

Stoichiometry helps us understand how to relate the amounts of reactants and products in a chemical reaction. It comes into play once we have a balanced chemical equation, which ensures the law of conservation of mass is followed. That means each type of atom must be the same on both sides of the equation.
In our example, the balanced chemical equation for sulfur dioxide reacting to form sulfuric acid is:

• \[ 2\,\mathrm{SO}_{2} + \mathrm{O}_{2} + 2\,\mathrm{H}_{2}O \rightarrow 2\,\mathrm{H}_{2}SO_{4} \]

To calculate the amount of sulfuric acid formed from sulfur dioxide, we look at the mole ratio provided by the balanced equation. Here, 2 moles of \(\mathrm{SO}_{2}\) yield 2 moles of \(\mathrm{H}_{2}SO_{4}\). This 1:1 ratio is a critical aspect of stoichiometry in determining the amount of products formed from a known amount of reactants.

Molar Mass Calculation

Understanding molar mass calculations is key to performing stoichiometric conversions. The molar mass is the weight of one mole of a given substance and is expressed in grams per mole (g/mol).
Let's begin with sulfur dioxide, \(\mathrm{SO}_{2}\). To calculate its molar mass, you need the atomic masses:

• Sulfur has a mass of 32.07 g/mol.
• Oxygen has a mass of 16.00 g/mol.

Therefore, the molar mass of \(\mathrm{SO}_{2}\) is:

• \[ 32.07 + 2 \times 16.00 = 64.07\, \text{g/mol} \]

Similarly, the molar mass of sulfuric acid, \(\mathrm{H}_{2}SO_{4}\), is found by adding:

• 2 hydrogen atoms (1.01 g/mol each)
• 1 sulfur atom (32.07 g/mol)
• 4 oxygen atoms (16.00 g/mol each)

The calculation is:

• \[ 2 \times 1.01 + 32.07 + 4 \times 16.00 = 98.09\, \text{g/mol} \]

These calculations are fundamental for converting between grams and moles, allowing us to apply stoichiometry effectively.

Sulfuric Acid Production

The formation of sulfuric acid is a significant reaction in the environment and industrial processes. It begins when sulfur dioxide ( \(\mathrm{SO}_{2}\)) reacts with oxygen and water in the atmosphere during specific weather conditions, contributing to the phenomenon known as acid rain.
In our problem, we want to calculate how much \(\mathrm{H}_{2}SO_{4}\) can be produced from 2.50 grams of \(\mathrm{SO}_{2}\).
Using its molar mass, we convert this to moles:

• \[ \text{Moles of } \mathrm{SO}_{2} = \frac{2.50\, \text{g}}{64.07\, \text{g/mol}} = 0.039\, \text{mol} \]

From stoichiometry, this means we also produce 0.039 moles of \(\mathrm{H}_{2}SO_{4}\), given the 1:1 ratio in the balanced equation. Finally, converting back to grams:

• \[ \text{Mass of } \mathrm{H}_{2}SO_{4} = 0.039\, \text{mol} \times 98.09\, \text{g/mol} = 3.82\, \text{g} \]

This calculation shows how we use stoichiometry and molar mass to determine the quantities of products formed in chemical reactions, highlighting why these concepts are necessary in both academic and real-world chemistry.