Problem 92 - General Chemistry

Question

How many milliliters of \(0.250 \mathrm{M} \mathrm{KMnO}_{4}\) are needed to react with \(3.36 \mathrm{~g}\) of iron(II) sulfate, \(\mathrm{FeSO}_{4}\) ? The reaction is as follows: $$ \begin{aligned} &10 \mathrm{FeSO}_{4}(a q)+2 \mathrm{KMnO}_{4}(a q)+8 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\ &5 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+2 \mathrm{MnSO}_{4}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$

Step-by-Step Solution

Answer

17.68 mL of 0.250 M KMnO₄ are needed.

Step 1: Find Molar Mass of FeSO₄

First, determine the molar mass of iron(II) sulfate (\( \mathrm{FeSO}_4 \)). The molar mass is the sum of the atomic masses: \( \mathrm{Fe} = 55.85 \), \( \mathrm{S} = 32.07 \), and \( \mathrm{O} = 16.00 \times 4 = 64.00 \). Therefore, the molar mass of \( \mathrm{FeSO}_4 \) is \( 55.85 + 32.07 + 64.00 = 151.92 \, \mathrm{g/mol} \).

Step 2: Calculate Moles of FeSO₄

Using the molar mass, calculate the moles of \( \mathrm{FeSO}_4 \) present in \( 3.36 \, \mathrm{g} \). Use the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \): \[ \text{moles of } \mathrm{FeSO}_4 = \frac{3.36 \, \mathrm{g}}{151.92 \, \mathrm{g/mol}} \approx 0.0221 \, \text{mol} \].

Step 3: Use Stoichiometry to Find Moles of KMnO₄

From the balanced chemical equation, \( 10 \) moles of \( \mathrm{FeSO}_4 \) react with \( 2 \) moles of \( \mathrm{KMnO}_4 \). Therefore, \( 0.0221 \, \text{mol} \) of \( \mathrm{FeSO}_4 \) will react with \( \frac{2}{10} \times 0.0221 \text{ mol} = 0.00442 \text{ mol} \) of \( \mathrm{KMnO}_4 \).

Step 4: Calculate Volume of KMnO₄ Solution Needed

The concentration of the \( \mathrm{KMnO}_4 \) solution is \( 0.250 \mathrm{M} \), which means \( 0.250 \text{ mol/L} \). Use the formula \( \text{volume} = \frac{\text{moles}}{\text{concentration}} \) to find the volume needed: \[ \text{volume} = \frac{0.00442 \, \text{mol}}{0.250 \, \text{mol/L}} = 0.01768 \, \text{L} \]. Convert this to milliliters: \( 0.01768 \, \text{L} \times 1000 \, \text{mL/L} = 17.68 \, \text{mL} \).

Key Concepts

Molar Mass

In chemistry, the molar mass of a compound is an essential concept that relates to the mass of one mole of that substance. To calculate it, you need to sum up the atomic masses of all the atoms in a molecule of the compound. For example, in the compound iron(II) sulfate, \( \mathrm{FeSO}_4 \), the molar mass is determined by adding up the atomic masses of iron (Fe), sulfur (S), and oxygen (O). Each element's atomic mass is specific, such as Fe being 55.85, S as 32.07, and O as 16.00. Since there are four oxygen atoms, their total contribution to the molar mass is 64.00. Therefore, the molar mass of \( \mathrm{FeSO}_4 \) is calculated as \( 55.85 + 32.07 + 64.00 = 151.92 \, \mathrm{g/mol} \). Understanding molar mass is crucial for converting between grams and moles, which is a common task in stoichiometry.

Chemical Reaction

A chemical reaction involves rearranging the molecules of one or more substances to form new substances. In the given exercise, we have a balanced chemical equation showing the reaction between iron(II) sulfate, potassium permanganate, and sulfuric acid:

• 10 \( \mathrm{FeSO}_4(aq) \)
• 2 \( \mathrm{KMnO}_4(aq) \)
• 8 \( \mathrm{H}_2 \mathrm{SO}_4(aq) \)

produce

• 5 \( \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3(aq) \)
• 2 \( \mathrm{MnSO}_4(aq) \)
• \(\mathrm{K}_2 \mathrm{SO}_4(aq) \)
• 8 \( \mathrm{H}_2 \mathrm{O}(l) \)

Each substance on the reactant side interacts and transforms into products, emphasizing the concept that the mass is conserved during the reaction. Balancing equations requires matching the number of each type of atom on both sides, ensuring that no atoms are lost or gained in the process. This balance allows us to use stoichiometry to determine the proportions of each substance needed or produced.

Solution Concentration

Solution concentration represents the amount of solute dissolved in a given volume of solvent, typically expressed in units of moles per liter (Molarity, M). In the exercise, the \(\mathrm{KMnO}_4 \) solution has a concentration of \(0.250 \, \mathrm{M}\). This indicates that there are \(0.250 \, \text{moles of } \mathrm{KMnO}_4\) in every liter of solution. Understanding solution concentration is crucial for performing dilution calculations and predicting how solutions will behave when mixed. It gives us a straightforward way to calculate how much of a solution is needed to react with a given amount of another substance. By knowing the moles required, and using the concentration (M), you can determine how much solution is needed for a particular reaction.

Volume Measurement

Volume measurement is vital in experiments and calculations, especially in chemical reactions involving solutions. It refers to the amount of space a substance occupies and is usually measured in liters (L) or milliliters (mL). For solutions, the volume needed is often calculated based on the moles of solute required and the concentration of the solution. In this case, you first determine the moles of \(\mathrm{KMnO}_4\) needed and then use its concentration to find out the volume. The formula used is \( \text{volume} = \frac{\text{moles}}{\text{concentration}} \), which in this exercise results in the calculation of \(0.01768 \, \text{L}\). Converting liters to milliliters by multiplying by 1000 gives \(17.68 \, \text{mL}\). Correct volume measurement is essential for ensuring the reaction proceeds as expected and for the correct amount of products to be formed.