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 g of \(\mathrm{SO}_{2}\) reacts with excess oxygen and water, how much \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in grams, is produced?

Short answer

The balanced chemical equation for the reaction is: \(\mathrm{SO_2} + \frac{1}{2}\mathrm{O_2} + \mathrm{H_2O} \rightarrow \mathrm{H_2SO_4}\). When 2.50 grams of \(\mathrm{SO_2}\) reacts with excess oxygen and water, approximately 3.83 grams of \(\mathrm{H_2SO_4}\) is produced.

Step-by-step solution

Write the balanced chemical equation

The unbalanced chemical equation for the reaction is: \[\mathrm{SO_2} + \mathrm{O_2} + \mathrm{H_2O} \rightarrow \mathrm{H_2SO_4}\] To balance the equation, we must make sure the number of atoms of each element on both sides is equal. In this case, we have: \[\mathrm{S} : 1 = 1, \mathrm{O} : 2 + 2 + 1 = 1 \times 4, \mathrm{H} : 2 = 2\] The balanced chemical equation is: \[\mathrm{SO_2} + \frac{1}{2}\mathrm{O_2} + \mathrm{H_2O} \rightarrow \mathrm{H_2SO_4}\]

Calculate the moles of sulfur dioxide

Given that 2.50 grams of \(\mathrm{SO}_{2}\) reacts, let's calculate the moles of \(\mathrm{SO}_{2}\): Molar mass of \(\mathrm{SO}_{2} = 32.1 \,(\mathrm{S}) + 2 \times 16 \,(\mathrm{O}) = 64.1 \, \mathrm{g/mol}\) \[moles \, of \, \mathrm{SO}_{2} = \frac{mass}{molar \, mass} = \frac{2.50 \, \mathrm{g}}{64.1 \, \mathrm{g/mol}} \approx 0.039 \, \mathrm{mol}\]

Determine the moles of produced sulfuric acid

From the balanced chemical equation in Step 1, we see that 1 mole of \(\mathrm{SO_2}\) produces 1 mole of \(\mathrm{H_2SO_4}\). Therefore, 0.039 moles of \(\mathrm{SO_2}\) will produce 0.039 moles of \(\mathrm{H_2SO_4}\) because there's excess oxygen and water.

Calculate the mass of produced sulfuric acid

Now we can calculate the mass of \(\mathrm{H_2SO_4}\) produced: Molar mass of \(\mathrm{H_2SO_4} = 2 \times1 \,(\mathrm{H}) + 32.1 \,(\mathrm{S}) + 4 \times 16 \,(\mathrm{O}) = 98.1 \, \mathrm{g/mol}\) \[mass \, of \, \mathrm{H_2SO_4} = moles \, of \, \mathrm{H_2SO_4} \times molar \, mass = 0.039 \, \mathrm{mol} \times 98.1 \, \mathrm{g/mol} \approx 3.83 \, \mathrm{g}\] So, about 3.83 grams of \(\mathrm{H_2SO_4}\) is produced when 2.50 grams of \(\mathrm{SO_2}\) reacts with excess oxygen and water.

Key concepts

Balanced Chemical Equation

The creation of a balanced chemical equation is pivotal in understanding reactions like the formation of acid rain. In essence, it serves as the recipe for the chemical process, illustrating how reactants combine to form products. Correctly balancing these equations ensures that the law of conservation of mass is respected, indicating that atoms are neither created nor destroyed during the reaction.

For the formation of sulfuric acid from sulfur dioxide, we write down all reactants and products and then adjust their coefficients to balance the atoms on both sides of the equation. This balancing act requires a bit of trial and error but there are systematic approaches, such as balancing atoms of elements that appear in the fewest compounds first or using the least common multiple to balance oxygen and hydrogen atoms last. Understanding the nuances of creating a balanced equation not only enhances one’s problem-solving skills but is fundamental in grasping the conservation principles inherent in chemistry.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantities of reactants and products in chemical reactions. It's based on the balanced chemical equation and the principle that reactants are converted to products in fixed ratios (stoichiometric coefficients) indicated by the equation.

In our example involving the formation of sulfuric acid, stoichiometry allows us to quantitatively relate the amount of sulfur dioxide used to the amount of sulfuric acid formed. Since the coefficients are in a 1:1 ratio, the moles of sulfur dioxide directly determine the moles of sulfuric acid produced. Grasping the core concepts of stoichiometry is essential for anyone looking to perform accurate chemical calculations or predict the outcomes of reactions under various conditions.

Molar Mass Calculation

Molar mass calculation is an essential skill to quantify the mass of substances involved in chemical reactions. The molar mass, which is the mass of one mole of a substance, is measured in grams per mole (g/mol). It can be determined by summing the molar masses of the individual elements in the compound, as provided by the periodic table.

For sulfur dioxide, this involves adding the molar mass of sulfur to twice the molar mass of oxygen. Similarly, the molar mass of sulfuric acid is obtained by summing the molar masses of two hydrogens, one sulfur, and four oxygens. Knowing the molar mass allows us to convert between the mass of a substance and the amount in moles, a key step in stoichiometric calculations as seen in the acid rain equation where grams of sulfur dioxide are converted to grams of sulfuric acid produced.