Problem 16
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?
Step-by-Step Solution
Verified 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.
1Step 1: 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:
\[\mathrm{SO_2} + \frac{1}{2}\mathrm{O_2} + \mathrm{H_2O} \rightarrow \mathrm{H_2SO_4}\]
2Step 2: 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 \(\mathrm{SO}_{2} = 32.1 \,(\mathrm{S}) + 2 \times 16 \,(\mathrm{O}) = 64.1 \, \mathrm{g/mol}\)
\[moles \, of \, \mathrm{SO}_{2} = \frac{mass}{molar \, mass} = \frac{2.50 \, \mathrm{g}}{64.1 \, \mathrm{g/mol}} \approx 0.039 \, \mathrm{mol}\]
3Step 3: 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.
4Step 4: 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 EquationStoichiometryMolar Mass Calculation
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.
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.
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.
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.
Other exercises in this chapter
Problem 14
Challenge Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride \(\left(\mathrm{TiCl}_{4}\ri
View solution Problem 15
One of the reactions used to inflate automobile air bags involves sodium azide \(\left(\mathrm{NaN}_{3}\right) : 2 \mathrm{NaN}_{3}(\mathrm{s}) \rightarrow 2 \m
View solution Problem 17
Explain why a balanced chemical equation is needed to solve a stoichiometric problem.
View solution Problem 18
List the four steps used in solving stoichiometric problems.
View solution