Question

Acid precipitation is the term generally used to describe rain or snow that is more acidic than it normally is. One cause of acid precipitation is the formation of sulfuric and nitric acids from various sulfur and nitrogen oxides produced in volcanic eruptions, forest fires, and thunderstorms. In a typical volcanic eruption, for example, \(3.50 \times 10^{8} \mathrm{kg} \mathrm{SO}_{2}\) may be produced. If this amount of \(\mathrm{SO}_{2}\) were converted to \(\mathrm{H}_{2} \mathrm{SO}_{4}\) according to the two-step process given below, how many kilograms of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) would be produced from such an eruption? $$ \begin{array}{c}{\mathrm{SO}_{2}+\frac{1}{2} \mathrm{O}_{2} \longrightarrow \mathrm{SO}_{3}} \\ {\mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}}\end{array} $$

Short answer

5.35 \times 10^8 \text{kg}

Step-by-step solution

- Calculate the moles of \( \mathrm{SO}_{2} \)

First, calculate the number of moles of \( \mathrm{SO}_{2} \). Use the molar mass of \( \mathrm{SO}_{2}, which is 64.07 \mathrm{g/mol} \). \[\text{Moles of SO}_{2} = \frac{3.50 \times 10^{8} \text{kg}}{64.07 \text{g/mol}} = \frac{3.50 \times 10^{11} \text{g}}{64.07 \text{g/mol}} \approx 5.46 \times 10^9 \text{mol}\]

- Determine the moles of \( \mathrm{H}_{2} \mathrm{SO}_{4} \)

According to the reactions given: \[\begin{array}{c}{\text{SO}_{2} + \frac{1}{2} \text{O}_{2} \rightarrow \text{SO}_{3}} \end{array} \] \[\begin{array}{c}{\text{SO}_{3} + \text{H}_{2}O \rightarrow \text{H}_{2}SO_{4}} \end{array} \] one mole of \text{SO}_{2} produces one mole of \text{H}_{2}SO_{4}. Thus, the number of moles of \text{H}_{2}SO_{4} produced is equal to the number of moles of \text{SO}_{2}: \[\text{Moles of H}_{2}SO_{4} = 5.46 \times 10^9 \text{mol}\]

- Calculate the mass of \( \mathrm{H}_{2} \mathrm{SO}_{4} \)

Finally, use the molar mass of \text{H}_{2}SO_{4} which is 98.09 \text{g/mol}, to find the mass produced. \[\text{Mass of H}_{2}SO_{4} = \text{Moles of H}_{2}SO_{4} \times \text{Molar Mass of H}_{2}SO_{4} \approx 5.46 \times 10^9 \text{mol} \times 98.09 \text{g/mol} = 5.35 \times 10^{11} \text{g} = 5.35 \times 10^8 \text{kg}\]

Key concepts

sulfuric acid formation

Sulfuric acid formation is a crucial part of understanding acid precipitation. Sulfuric acid \(\text{H}_{2}\text{SO}_{4}\) is a strong acid formed through various chemical reactions. Typically, sulfur dioxide \(\text{SO}_{2}\) produced from volcanic eruptions, forest fires, or other natural phenomena reacts with oxygen \(\text{O}_{2}\) and water \(\text{H}_{2}\text{O}\). This involves a two-step process:

• First, sulfur dioxide \(\text{SO}_{2}\) reacts with oxygen \(\text{O}_{2}\) to form sulfur trioxide \(\text{SO}_{3}\).
• Second, sulfur trioxide \(\text{SO}_{3}\) reacts with water \(\text{H}_{2}\text{O}\) to form sulfuric acid \(\text{H}_{2}\text{SO}_{4}\).

The overall reaction can be written as: \(\text{SO}_{2} + \frac{1}{2} \text{O}_{2} \rightarrow \text{SO}_{3}\) and \(\text{SO}_{3} + \text{H}_{2}\text{O} \rightarrow \text{H}_{2}\text{SO}_{4}\). Each step is crucial for the complete transformation.

chemical reactions

Chemical reactions involve the transformation of reactants into products. In our case, the main reactions are:
\(\text{SO}_{2} + \frac{1}{2} \text{O}_{2} \rightarrow \text{SO}_{3}\)
\(\text{SO}_{3} + \text{H}_{2}\text{O} \rightarrow \text{H}_{2}\text{SO}_{4}\)

These reactions show the conversion of sulfur dioxide to sulfuric acid.

• The first reaction converts \(\text{SO}_{2}\) to sulfur trioxide (\text{SO}_{3}). This is oxidation, where oxygen is added.
• The second reaction combines sulfur trioxide (\text{SO}_{3}) with water (\text{H}_{2}\text{O}) to form sulfuric acid (\text{H}_{2}\text{SO}_{4}). Here, \(\text{SO}_{3}\) and \(\text{H}_{2}\text{O}\) are reactants forming a single product, \(\text{H}_{2}\text{SO}_{4}\).

Understanding these chemical reactions helps in grasping how acid precipitation is formed and its environmental effects.

mole calculations

Mole calculations let us quantify substances in chemical reactions. The mole concept helps in converting mass to the number of particles (atoms, molecules).

Here, we calculated moles of \(\text{SO}_{2}\) using its molar mass:
\(\text{Moles of SO}_{2} = \frac{3.50 \times 10^{8} \text{kg}}{64.07 \text{g/mol}} = \frac{3.50 \times 10^{11} \text{g}}{64.07 \text{g/mol}} = 5.46 \times 10^9 \text{mol}\)

Converting mass to moles involves dividing the mass by the molar mass. Here, we start with 350 million kg of \(\text{SO}_{2}\), convert to grams (1 kg = 1000 g), and then use the molar mass (64.07 g/mol).

This calculation helps in determining how much sulfuric acid can be formed.

stoichiometry

Stoichiometry deals with the quantitative relationships between reactants and products in chemical reactions. It allows us to predict how much product we can get from a given amount of reactant.

For sulfuric acid formation:

• First, calculate the moles of \(\text{SO}_{2}\) which is approximately 5.46 x 10\text{9 mol}.
• From the balanced reactions, we know that 1 mole of \(\text{SO}_{2}\) produces 1 mole of \(\text{H}_{2}\text{SO}_{4}\). Thus, 5.46 x 10\text{9 mol} of \(\text{SO}_{2}\) will produce an equal number of moles of \(\text{H}_{2}\text{SO}_{4}\).
• Finally, convert moles of \(\text{H}_{2}\text{SO}_{4}\) to mass using its molar mass (98.09 g/mol): \(\text{Mass} = \text{Moles} \times \text{Molar Mass}\). Hence, \(\text{5.46 x 10^9 mol \times 98.09 g/mol = 5.35 x 10^8 kg}\).

Stoichiometry ensures precise calculations in chemical reactions, helping us know exactly how much product is formed.