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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 ( KMnOâ‚„ ), oxalic acid ( Hâ‚‚Câ‚‚Oâ‚„ ), and sulfuric acid ( Hâ‚‚SOâ‚„ ). These reactants transform into several products including manganese sulfate ( MnSOâ‚„ ), carbon dioxide ( COâ‚‚ ), water ( Hâ‚‚O ), and potassium sulfate ( Kâ‚‚SOâ‚„ ). 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 KMnOâ‚„ react with 5 moles of Hâ‚‚Câ‚‚Oâ‚„, reflecting the 2:5 stoichiometric ratio. This ratio is crucial for determining how much KMnOâ‚„ is needed to completely react with a given amount of oxalic acid.

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 (Hâ‚‚Câ‚‚Oâ‚„), 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 Hâ‚‚Câ‚‚Oâ‚„ 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:\[\text{Moles of } Hâ‚‚Câ‚‚Oâ‚„ = \frac{3.225 \, \mathrm{g}}{90.03 \, \mathrm{g/mol}} = 0.03582 \, \mathrm{mol}.\]This conversion to moles is the first step in applying stoichiometry for the reaction.

Solution concentration, 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 KMnOâ‚„ solution is 0.250 M.
Once the moles of KMnOâ‚„ 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.