Problem 99 - Chemistry The Molecular Science

Question

Uranium(VI) oxide reacts with bromine trifluoride to give uranium(IV) fluoride, an important step in the purification of uranium ore. \(6 \mathrm{UO}_{3}(\mathrm{~s})+8 \mathrm{BrF}_{3}(\ell) \longrightarrow 6 \mathrm{UF}_{4}(\mathrm{~s})+4 \mathrm{Br}_{2}(\ell)+9 \mathrm{O}_{2}(\mathrm{~g})\) You begin with \(365 \mathrm{~g}\) each of \(\mathrm{UO}_{3}\) and \(\mathrm{BrF}_{3}\); determine the maximum yield, in grams, of \(\mathrm{UF}_{4}\).

Step-by-Step Solution

Answer

The maximum yield of \(\mathrm{UF}_4\) is 400.61 g.

Step 1: Molar Mass Calculation

Calculate the molar masses of \(\mathrm{UO}_3, \mathrm{BrF}_3,\) and \(\mathrm{UF}_4\) using their atomic masses. \(\mathrm{UO}_3\) has a molar mass of 286.03 g/mol (U = 238.03, O = 16.00). \(\mathrm{BrF}_3\) has a molar mass of 136.90 g/mol (Br = 79.90, F = 19.00). \(\mathrm{UF}_4\) has a molar mass of 314.03 g/mol (U = 238.03, F = 19.00).

Step 2: Calculate Moles of Reactants

Determine the number of moles for each reactant. For \(\mathrm{UO}_3\), \(\frac{365 \text{ g}}{286.03 \text{ g/mol}} = 1.276 \text{ mol}\). For \(\mathrm{BrF}_3\), \(\frac{365 \text{ g}}{136.90 \text{ g/mol}} = 2.666 \text{ mol}\).

Step 3: Identify the Limiting Reactant

Use stoichiometry to find which reactant is limiting. The balanced equation shows that 6 moles of \(\mathrm{UO}_3\) react with 8 moles of \(\mathrm{BrF}_3\). Set up the ratio: \(\frac{1.276 / 6}{2.666 / 8}\). \(\mathrm{UO}_3\) is the limiting reactant because the ratio \(0.213 > 0.167\).

Step 4: Determine Maximum Yield of Product

Using the limiting reactant \(\mathrm{UO}_3\), find the theoretical yield of \(\mathrm{UF}_4\). Since 6 moles of \(\mathrm{UO}_3\) produce 6 moles of \(\mathrm{UF}_4\), 1.276 moles of \(\mathrm{UO}_3\) will produce 1.276 moles of \(\mathrm{UF}_4\).

Step 5: Convert Moles of \(\mathrm{UF}_4\) to Grams

Convert the moles of \(\mathrm{UF}_4\) to grams. Multiply the moles by the molar mass of \(\mathrm{UF}_4\): \(1.276 \text{ mol} \times 314.03 \text{ g/mol} = 400.61 \text{ g}\).

Key Concepts

Molar Mass Calculation

The calculation of molar mass is a fundamental step in stoichiometry. It provides us with essential information to understand how much of each substance is involved in a chemical reaction. Molar mass is defined as the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is derived from the sum of the atomic masses of all the atoms present in the compound.

For example, to calculate the molar mass of uranium(VI) oxide ( UO₃ ), we add the atomic mass of uranium ( 238.03 g/mol ) and three times the atomic mass of oxygen ( 16.00 g/mol ). This gives a molar mass of 286.03 g/mol for UO₃ . Similarly, the molar mass for bromine trifluoride ( BrF₃ ) is determined by summing the atomic mass of bromine ( 79.90 g/mol ) and three times the atomic mass of fluorine ( 19.00 g/mol ), totaling 136.90 g/mol .

• Uranium (VI) oxide ( UO₃ ): 286.03 g/mol
• Bromine trifluoride ( BrF₃ ): 136.90 g/mol
• Uranium (IV) fluoride ( UF₄ ): 314.03 g/mol

Understanding these values is crucial for converting between grams of a substance and the amount in moles, setting the stage for further stoichiometric calculations.

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is completely consumed first, stopping the reaction from proceeding further. It determines the amount of product that can be formed. To identify the limiting reactant, we rely on the balanced chemical equation and the amount of moles of each reactant available.

Consider the balanced equation for the reaction of UO₃ and BrF₃ :

• 6 mol UO₃ react with 8 mol BrF₃

Let's calculate the mole ratio using our reactants: - UO₃ : 1.276 mol - BrF₃ : 2.666 mol
Checking the proportion, UO₃ becomes the limiting reactant because its ratio of available moles compared to the equation-driven proportion is smaller. The limiting reactant is a key player in determining how much product will form because once it's used up, the reaction cannot proceed further.

Theoretical Yield

Theoretical yield is the maximum amount of product expected from a reaction, calculated based on stoichiometry and the limiting reactant. It assumes perfect conditions, where all of the limiting reactant converts into the desired product.

Using the example of UO₃ and BrF₃ , we determined that UO₃ is the limiting reactant. According to the balanced reaction, each mole of UO₃ will generate one mole of UF₄ . Hence, 1.276 moles of UO₃ will likewise produce 1.276 moles of UF₄ .

To find the theoretical yield in grams, we convert the moles of UF₄ to grams using its molar mass:

• Theoretical yield of UF₄ : 1.276 mol × 314.03 g/mol = 400.61 g

This calculation provides an ideal maximum yield, helping chemists understand the efficiency of the reaction under given conditions and guiding practical expectations in laboratory settings.