Problem 14 - Chemical Principles: The Quest for Insight

Question

Small amounts of chlorine gas can be generated in the laboratory from the reaction of manganese(IV) oxide with hydrochloric acid: \(4 \mathrm{HCl}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{~s})\) \(\rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{MnCl}_{2}(\mathrm{~s})+\mathrm{Cl}_{2}(\mathrm{~g})\). (a) What mass of \(\mathrm{Cl}_{2}\) can be produced from \(42.7 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\) with an excess of HCl(aq)? (b) What volume of chlorine gas (of density \(3.17 \mathrm{~g} \cdot \mathrm{L}^{-1}\) ) will be produced from the reaction of \(300 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) with an excess of \(\mathrm{MnO}_{2}\) ? (c) Suppose only \(150 \mathrm{~mL}\) of chlorine was produced in the reaction in (b). What is the percentage yield of the reaction?

Step-by-Step Solution

Answer

Part (a) 34.87 g of Cl2, part (b) 16.8 mL of Cl2, and part (c) the calculation shows more than 100% yield, which is not possible and suggests an error.

Step 1: Calculate Molar Mass of MnO2

To find out how much chlorine gas is produced from MnO2, calculate the molar mass by summing the atomic masses of its constituent elements. The atomic masses are: Mn = 54.94 g/mol, O = 16.00 g/mol. Thus, MnO2 has a molar mass of 54.94 g/mol + (2 × 16.00 g/mol) = 86.94 g/mol.

Step 2: Determine Moles of MnO2

Using the molar mass from Step 1, convert the mass of MnO2 to moles: 42.7 g MnO2 × (1 mol MnO2/86.94 g MnO2) = 0.491 moles of MnO2.

Step 3: Find Moles of Cl2 Produced

The stoichiometry of the reaction shows that 1 mole of MnO2 produces 1 mole of Cl2. Hence, 0.491 moles of MnO2 will produce 0.491 moles of Cl2.

Step 4: Calculate Mass of Cl2 Produced

Now convert the moles of Cl2 to mass using the molar mass of Cl2 (71.0 g/mol): 0.491 moles Cl2 × 71.0 g/mol = 34.87 g of Cl2.

Step 5: Determine Moles of HCl Reacted

300 mL of 0.100 M HCl contains 0.0300 L × 0.100 mol/L = 0.00300 moles of HCl. Since the reaction requires 4 moles of HCl for every 1 mole of Cl2 produced, the limiting reagent is HCl.

Step 6: Find Moles of Cl2 from HCl

Using the stoichiometry, 0.00300 moles of HCl would produce 0.00300/4 = 0.000750 moles of Cl2 when MnO2 is in excess.

Step 7: Calculate Volume of Cl2 at Given Density

Using the density of Cl2, convert the mass of Cl2 to volume: Mass = Density × Volume. Volume is thus Mass/Density, and therefore, the volume of Cl2 produced is (0.000750 moles × 71.0 g/mol) / 3.17 g/L = 0.0168 L or 16.8 mL of Cl2.

Step 8: Calculate the Percentage Yield

Percentage yield is calculated using the actual yield (150 mL given) and the theoretical yield (16.8 mL from Step 7). Percentage yield = (Actual Yield / Theoretical Yield) × 100% = (150 mL / 16.8 mL) × 100% = 892.86%. Since a yield above 100% is not possible, it indicates there might be a mistake in the actual yield or in the calculation.

Key Concepts

Molar Mass Calculation

Understanding the molar mass of a substance is crucial for stoichiometry problems. Molar mass is defined as the mass of one mole of a given substance and is measured in grams per mole (g/mol). It is equivalent to the sum of the atomic masses of each element in a compound, taken from the periodic table, and multiplied by the number of atoms of that element in the formula.

For example, manganese(IV) oxide (\textbf{MnO}\(_2\)) has a molar mass calculated by adding the atomic mass of manganese (Mn), which is 54.94 g/mol, to twice the atomic mass of oxygen (O), which is 16.00 g/mol, because there are two oxygen atoms. This gives us: \(54.94 \text{ g/mol } + 2 \times 16.00 \text{ g/mol } = 86.94 \text{ g/mol}\).

Mole-to-Mass Conversion

Once you have the molar mass, converting between moles and grams becomes straightforward. To convert grams to moles, you divide the mass of the substance by its molar mass. Conversely, to go from moles to grams, multiply the number of moles by the molar mass.

In stoichiometry problems, this mole-to-mass conversion allows us to start with a known mass of a reactant or product and find out how many moles we are dealing with, which in turn lets us use the balanced chemical equation to determine the amounts of other substances involved in the reaction.

Limiting Reagent Determination

Identifying the limiting reagent is a critical step in stoichiometry. The limiting reagent is the reactant that will be completely used up first and thus limits the extent of the reaction. To find the limiting reagent, compare the mole ratio of each reactant in the balanced equation to the mole ratio of the reactants available.

If a reactant is present in a lower mole ratio than required, it is the limiting reagent. Knowing which reactant is limiting allows you to accurately predict how much product can be formed. In our example, HCl is identified as the limiting reagent, since it restricts the quantity of chlorine gas that can be generated.

Reaction Yield Calculation

The yield of a reaction refers to the amount of product obtained. The theoretical yield is the maximum amount of product expected based on stoichiometry, assuming complete conversion of limiting reagent. The actual yield is the measured amount of product obtained from the experiment.

The percent yield is a comparison of the two: \(\text{Percent Yield} = (\text{Actual Yield} / \text{Theoretical Yield}) \times 100\text{%}\). It's a measure of the efficiency of a reaction. A yield above 100% typically indicates an error in the experiment or calculations--no reaction can produce more product than the stoichiometry predicts.

Gas Density and Volume Relationship

In gas stoichiometry problems, understanding the relationship between gas density and volume is essential. Density is the mass of the gas per unit volume, expressed as g/L. To find the volume from the mass of a gas, rearrange the equation for density: \(\text{Volume} = \text{Mass} / \text{Density}\).

This formula is particularly helpful when dealing with gases at standard temperature and pressure (STP) where the densities can be used to convert between mass and volume directly. However, for gases not at STP, ideal gas law calculations might be required to account for temperature and pressure effects.