Question

When combined, aqueous solutions of sulfuric acid and potassium hydroxide react to form water and aqueous potassium sulfate according to the following equation (unbalanced): $$ \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{KOH}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) $$ Determine what mass of water is produced when a beaker containing \(100.0 \mathrm{~g} \mathrm{H}_{2} \mathrm{SO}_{4}\) dissolved in \(250 \mathrm{~mL}\) water is added to a larger beaker containing \(100.0 \mathrm{~g}\) KOH dissolved in \(225 \mathrm{~mL}\) water. Determine the mass amounts of each substance (other than water) present in the large beaker when the reaction is complete.

Short answer

Begin by balancing the equation.

Step-by-step solution

Write the Balanced Chemical Equation

First, to solve this problem, we must balance the chemical equation:\[ \mathrm{H}_{2} \mathrm{SO}_{4}(aq) + 2 \mathrm{KOH}(aq) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{K}_{2} \mathrm{SO}_{4}(aq) \]

Key concepts

Balanced Chemical Equation

A balanced chemical equation is essential for understanding how substances react with each other in a chemical reaction. In a chemical equation, reactants are transformed into products. To balance the equation, we ensure that there is the same number of each type of atom on both sides of the equation.
For the reaction between sulfuric acid and potassium hydroxide, the unbalanced equation is:\[\mathrm{H}_{2}\mathrm{SO}_{4}(aq) + \mathrm{KOH}(aq) \rightarrow \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{K}_{2}\mathrm{SO}_{4}(aq)\]
This equation is then balanced by adjusting the coefficients. We need 2 moles of KOH for every mole of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), so each type of atom equals on both sides:\[\mathrm{H}_{2}\mathrm{SO}_{4}(aq) + 2 \mathrm{KOH}(aq) \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{K}_{2}\mathrm{SO}_{4}(aq)\]
This balanced equation tells us that one molecule of sulfuric acid reacts with two molecules of potassium hydroxide to form two molecules of water and one molecule of potassium sulfate.

Mass Calculation

Mass calculation in stoichiometry allows us to determine the mass of substances involved in a chemical reaction. When you have a balanced chemical equation, you can use the concept of molar mass to convert between moles and grams.
We have to first calculate the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) and \(\mathrm{KOH}\) present in their respective solutions. The molar mass of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is approximately 98.08 g/mol, and for \(\mathrm{KOH}\), it is around 56.11 g/mol.
To find the moles:

• For \(100.0 \, \text{g}\) of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), moles are computed as \(\frac{100.0}{98.08}\approx 1.02\, \text{mol}\).
• For \(100.0 \, \text{g}\) of \(\mathrm{KOH}\), moles are \(\frac{100.0}{56.11}\approx 1.78\, \text{mol}\).

Using the balanced equation, we understand that 1 mole of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) reacts with 2 moles of \(\mathrm{KOH}\). So, \(\mathrm{H}_{2}\mathrm{SO}_{4}\) will be the limiting reagent due to its smaller amount relative to the needed ratio, and it determines the amount of product formed.
The balanced equation shows that 1 mole of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) produces 2 moles of \(\mathrm{H}_{2}\mathrm{O}\). Hence, 1.02 moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) will produce:\[1.02 \times 2 = 2.04 \, \text{moles of } \mathrm{H}_{2}\mathrm{O}\]
The molar mass of \(\mathrm{H}_{2}\mathrm{O}\) is 18.02 g/mol, so the mass of water produced is:\[2.04 \times 18.02 = 36.76 \, \text{g}\]

Chemical Reaction

A chemical reaction involves the transformation of reactants into products. It involves breaking old bonds and forming new ones, which results in new substances.
In our example, the chemical reaction occurs between aqueous sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)) and potassium hydroxide (\(\mathrm{KOH}\)). As they combine:

• Hydrogen ions from \(\mathrm{H}_{2}\mathrm{SO}_{4}\) combine with hydroxide ions from \(\mathrm{KOH}\) to form water (\(\mathrm{H}_{2}\mathrm{O}\)).
• The remaining sulfate ions combine with potassium ions to create potassium sulfate (\(\mathrm{K}_{2}\mathrm{SO}_{4}\)).

Each of these processes occurs simultaneously as the reactants dissolve in water. The reaction is an example of a double displacement reaction, where the ions of the reactants exchange partners to form new compounds. This type of reaction is common in solutions, especially with acids and bases.
Completing the reaction forms two distinct products. Water is formed as a liquid, while the other product, potassium sulfate, remains dissolved in solution. Understanding this process helps predict product formation and the quantities needed or produced in practical applications through mass calculations.