Question

\(\mathrm{KMnO}_{4}\) reacts with oxalic acid according to the equation $$ 2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O} $$ Here, \(20 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{KMnO}_{4}\) is equivalent to : (a) \(120 \mathrm{~mL}\) of \(0.25 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (b) \(150 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (c) \(25 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (d) \(50 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)

Short answer

The equivalent oxalic acid volume and molarity for 20 mL of 0.1 M KMnO4 is 25 mL of 0.20 M H2C2O4.

Step-by-step solution

Understanding the Stoichiometry

Identify the molar ratio between KMnO4 and oxalic acid (H2C2O4) from the balanced chemical equation. The molar ratio is 2 moles of KMnO4 to 5 moles of H2C2O4.

Calculate the Moles of KMnO4

Calculate the moles of KMnO4 used in the reaction. Use the volume and molarity given: Moles = Molarity x Volume = 0.1 M x 20 mL = 0.002 moles of KMnO4.

Calculate the Moles of Oxalic Acid

Use the stoichiometric ratio from Step 1 to find the moles of oxalic acid that react with the KMnO4. From the ratio, 5 moles of H2C2O4 react with 2 moles of KMnO4, hence the moles of H2C2O4 = (5/2) x 0.002 moles of KMnO4 = 0.005 moles of H2C2O4.

Compare with Given Options

Check which option gives 0.005 moles of H2C2O4 when using the respective volume and molarity. For each option, calculate the moles of H2C2O4: (a) 0.25 M x 120 mL = 0.03 moles, (b) 0.10 M x 150 mL = 0.015 moles, (c) 0.20 M x 25 mL = 0.005 moles, (d) 0.20 M x 50 mL = 0.01 moles.

Select the Correct Option

The correct option is the one where the moles of oxalic acid are 0.005 moles, which is option (c) 25 mL of 0.20 M H2C2O4.

Key concepts

Chemical Reaction Calculations

Chemical reaction calculations are essential for understanding how different substances interact in a chemical reaction. The calculations are based on the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. This law implies that the number of atoms of each element must remain constant before and after a chemical reaction, leading to the need for a balanced chemical equation.

When performing these calculations, one must consider the stoichiometry of the reaction, which involves the quantitative relationship between reactants and products. In our exercise, the balanced equation provided the ratio of reactants to products necessary for the reaction to occur without any excess of any reactant. For example, to find the amount of oxalic acid that reacts with a given amount of KMnO4, we use the stoichiometric coefficients from this balanced equation.

Step-by-step calculation helps us convert quantities from one form to another, such as converting volume and molarity into moles. Consistency in units is crucial; the common practice is to convert all volumes into liters and use the SI unit for moles, ensuring a standardized approach for calculations. With correct chemical reaction calculations, students can reliably predict the outcomes of reactions in lab or industry settings.

Molarity and Concentration

Molarity, often denoted by the letter M, is a measure of the concentration of a solute in a solution. It's defined as the number of moles of solute per liter of solution. This concentration measurement is extremely useful in chemistry because it allows scientists to predict how many particles of solute are present in a given volume of solution, which is essential for stoichiometry calculations.

To calculate the molarity of a substance, divide the number of moles of the solute by the volume of the solution in liters. This is shown in the formula:
\[ \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
Using molarity, chemists can quickly relate the volume of a solution to the amount of reactant or product in a chemical reaction. This concept is demonstrated in our textbook exercise where we calculate the number of moles of KMnO4 and compare different volumes and molarities of oxalic acid to find the equivalent solution. Understanding molarity is fundamental for students to grasp dilutions, compound formations, and reaction stoichiometry.

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry that ensures the law of conservation of mass is upheld. A balanced equation has an equal number of each type of atom on both sides of the reaction. It's vital for stoichiometry as it provides the ratio of reactants to products, known as the stoichiometric coefficients, which determine the proportions necessary for a reaction to occur completely.

When balancing equations, it's best to start with the most complex molecule and proceed to balance atoms that appear in fewer compounds. Sometimes, to achieve balance, you might need to use fractional coefficients, which are later multiplied to give whole numbers. In our exercise, the equation was already balanced; hence, we directly used it for mole calculations.

Properly balancing equations prevents errors in subsequent calculations and is the cornerstone of quantitative chemical analysis. It’s the first step before any chemical reaction calculations can be done, as seen when we needed the correct stoichiometric ratios to solve our textbook problem. Students must master balancing to accurately predict the amounts of substances consumed and produced in reactions.