Problem 37 - Chemistry: Principles and Reactions

Question

37\. Iodine can be prepared by allowing an aqueous solution of hydrogen iodide to react with manganese dioxide, \(\mathrm{MnO}_{2}\). The reaction is $$2 \mathrm{I}^{-}(a q)+4 \mathrm{H}^{+}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}(s)$$ If an excess of hydrogen iodide is added to \(0.200 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\), how many grams of iodine are obtained, assuming \(100 \%\) yield?

Step-by-Step Solution

Answer

Answer: The mass of iodine obtained is 0.583 g.

Step 1: Convert the mass of MnO2 to moles

To perform stoichiometry calculations, we need to convert the mass of the reactant, manganese dioxide (MnO2), into moles. We do this by dividing the mass by the molar mass of MnO2: Molar mass of MnO2 = 54.94 (Mn) + 2 * 16.00 (O) = 86.94 g/mol Moles of MnO2 = (mass of MnO2) / (molar mass of MnO2) Moles of MnO2 = 0.200 g / 86.94 g/mol = 0.00230 mol

Step 2: Use the stoichiometry of the reaction to determine moles of I2

From the balanced chemical equation, we can see that one mole of MnO2 reacts to produce one mole of I2. Therefore, if we have 0.00230 moles of MnO2, we will also have 0.00230 moles of I2 produced.

Step 3: Convert moles of I2 to mass

Now we need to convert moles of iodine (I2) back into mass. We do this by multiplying the moles by the molar mass of I2: Molar mass of I2 = 2 * 126.90 g/mol = 253.80 g/mol Mass of I2 = (moles of I2) * (molar mass of I2) Mass of I2 = 0.00230 mol * 253.80 g/mol = 0.583 g

Step 4: Final answer

Assuming a 100% yield, the mass of iodine obtained from a 0.200 g sample of manganese dioxide (MnO2) would be 0.583 g of I2.

Key Concepts

Chemical Reactions

When substances interact to form new compounds, a chemical reaction occurs. This process involves the breaking of old chemical bonds and the formation of new ones. In our case, hydrogen iodide reacts with manganese dioxide to produce iodine. Each reaction can be understood as a sequence of steps where reactants turn into products.
Key points about chemical reactions:

• They often involve a color change, gas formation, precipitate formation, or energy change.
• In a balanced chemical reaction, the mass and number of atoms are conserved.

For example, in the given equation, hydrogen iodide (HI) and manganese dioxide (\(\mathrm{MnO}_{2}\)) are the reactants, while iodine (\(\mathrm{I}_2\)) is the product formed as a solid. This reaction demonstrates how specific reactants in the right conditions can form desired products such as iodine.

Molar Mass Calculation

Calculating molar mass is essential in stoichiometry as it allows us to convert grams to moles. The molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol). To calculate it, you sum the atomic masses of all atoms in the molecule. For instance, in our problem:

• The molar mass of \(\mathrm{MnO}_2\) is calculated by adding 54.94 (manganese) + 2 \(\times\) 16.00 (oxygen) = 86.94 g/mol.
• For iodine \(\mathrm{I}_2\), it's twice the atomic mass of iodine: 2 \(\times\) 126.90 g/mol = 253.80 g/mol.

These calculations let us move between measurements in grams and the abstract concept of moles, which is vital for determining proportions of substances in reactions.

Chemical Equation Balancing

Balancing chemical equations ensures that the same number of each type of atom appears on both sides of the equation. This maintains the law of conservation of mass, stating that mass cannot be created or destroyed in a chemical reaction. To balance an equation:

• Identify each element in the given reactants and products.
• Adjust coefficients, not subscripts, to ensure equal numbers of each type of atom on both sides.

In our reaction of interest: \[2 \mathrm{I}^- (aq) + 4 \mathrm{H}^+ (aq) + \mathrm{MnO}_{2} (s) \longrightarrow \mathrm{Mn}^{2+} (aq) + 2 \mathrm{H}_{2}\mathrm{O} + \mathrm{I}_{2} (s)\]The coefficients indicate the proportions in which molecules react and are formed. Here, 2 moles of iodide ions and 4 moles of hydrogen ions react with 1 mole of manganese dioxide to produce 1 mole of manganese ion, 2 moles of water, and 1 mole of iodine. Balancing equations is crucial for correctly predicting the quantities of reactants and products in a reaction.