Question

Potassium permanganate \(\left(\mathrm{KMnO}_{4}\right)\) reacts with oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) in aqueous sulfuric acid according to the following equation: $$ \begin{gathered} 2 \mathrm{KMnO}_{4}(a q)+5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\ 2 \mathrm{MnSO}_{4}(a q)+10 \mathrm{CO}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \end{gathered} $$ How many milliliters of a \(0.250 \mathrm{M} \mathrm{KMnO}_{4}\) solution are needed to react completely with \(3.225 \mathrm{~g}\) of oxalic acid?

Short answer

57.32 mL of 0.250 M KMnO₄ is needed.

Step-by-step solution

Calculate Moles of Oxalic Acid (H₂C₂O₄)

First, find the molar mass of oxalic acid \( \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \). It is calculated as: \( 2(1.01) + 2(12.01) + 4(16.00) = 90.03 \, \mathrm{g/mol} \).Now, calculate the moles of oxalic acid in \( 3.225 \, \mathrm{g} \):\[\text{Moles of } \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} = \frac{3.225 \, \mathrm{g}}{90.03 \, \mathrm{g/mol}} = 0.03582 \, \mathrm{mol}.\]

Use Stoichiometry to Find Moles of KMnO₄

From the balanced chemical equation:\[2 \mathrm{KMnO}_{4} + 5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightarrow \cdots\]The mole ratio of \( \mathrm{KMnO}_{4} \) to \( \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \) is \( 2:5 \).Calculate moles of \( \mathrm{KMnO}_{4} \) needed for 0.03582 mol of \( \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \):\[\text{Moles of } \mathrm{KMnO}_{4} = \frac{2}{5} \times 0.03582 \, \mathrm{mol} = 0.01433 \, \mathrm{mol}.\]

Calculate Volume of KMnO₄ Solution

To find the volume of \( 0.250 \, \mathrm{M} \) \( \mathrm{KMnO}_{4} \) solution needed:\[\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}\]Rearranging gives \( \text{Volume (L)} = \frac{\text{moles}}{\text{Molarity}} \).\[\text{Volume} = \frac{0.01433 \, \mathrm{mol}}{0.250 \, \mathrm{M}} = 0.05732 \, \mathrm{L}\]Convert liters to milliliters:\[0.05732 \, \mathrm{L} \times 1000 = 57.32 \, \mathrm{mL}.\]

Key concepts

Chemical Reactions

Chemical reactions are the heart of chemistry and involve the transformation of substances through breaking and forming bonds. In our exercise, the chemical reaction encompasses potassium permanganate ( KMnO₄ ), oxalic acid ( H₂C₂O₄ ), and sulfuric acid ( H₂SO₄ ). These reactants transform into several products including manganese sulfate ( MnSO₄ ), carbon dioxide ( CO₂ ), water ( H₂O ), and potassium sulfate ( K₂SO₄ ). This specific reaction is a redox (oxidation-reduction) reaction where electrons are transferred between species.
The balanced equation provides valuable information about the stoichiometry of the reaction. Stoichiometry is simply a way to quantify relationships between reactants and products in a chemical equation. In our case, the balanced equation tells us that 2 moles of KMnO₄ react with 5 moles of H₂C₂O₄, reflecting the 2:5 stoichiometric ratio. This ratio is crucial for determining how much KMnO₄ is needed to completely react with a given amount of oxalic acid.

Molar Mass Calculation

Calculating molar mass is an essential skill in chemistry, aiding in the conversion from grams to moles, and vice versa. Molar mass is defined as the mass of one mole of a substance, measured in grams per mole (g/mol). For determining the molar mass of oxalic acid (H₂C₂O₄), we sum up the atomic masses of its constituent atoms:

• Hydrogen (H) with 2 atoms has a total mass of 2 x 1.01 g/mol.
• Carbon (C) with 2 atoms gives 2 x 12.01 g/mol.
• Oxygen (O) with 4 atoms results in 4 x 16.00 g/mol.

The total molar mass for H₂C₂O₄ is 90.03 g/mol.
With the given mass of 3.225 grams of oxalic acid, we then calculate moles by dividing the mass by the molar mass:\[\text{Moles of } H₂C₂O₄ = \frac{3.225 \, \mathrm{g}}{90.03 \, \mathrm{g/mol}} = 0.03582 \, \mathrm{mol}.\]This conversion to moles is the first step in applying stoichiometry for the reaction.

Solution Concentration

Solution concentration, often expressed as molarity (M), is the measure of how much solute is dissolved in a given volume of solution. It is expressed as the number of moles of solute per liter of solution (\frac{mol}{L}). In the exercise at hand, we know the concentration of the KMnO₄ solution is 0.250 M.
Once the moles of KMnO₄ required for reaction are calculated (0.01433 mol), we use the formula for molarity to find out the volume of the solution needed:\[ \text{Volume (L)} = \frac{\text{moles}}{\text{Molarity}} \]Plugging in our values:\[ \text{Volume} = \frac{0.01433 \, \mathrm{mol}}{0.250 \, \mathrm{M}} = 0.05732 \, \mathrm{L} \]Since we often measure volumes in milliliters, we convert liters to milliliters by multiplying by 1000:\[ 0.05732 \, \mathrm{L} \times 1000 = 57.32 \, \mathrm{mL}. \]This final volume is essential for precise chemical experiments, ensuring enough reactant is available without excess.