Problem 66 - Chemistry The Science in Context

Question

The uranium minerals found in nature must be refined and enriched in \(^{235} \mathrm{U}\) before the uranium can be used as a fuel in nuclear reactors. One procedure for enriching uranium relies on the reaction of \(\mathrm{UO}_{2}\) with HF to form UF \(_{4}\), which is then converted into UF \(_{6}\) by reaction with fluorine: $$\begin{array}{c}\mathrm{UO}_{2}(g)+4 \mathrm{HF}(a q) \rightarrow \mathrm{UF}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \\\\\mathrm{UF}_{4}(g)+\mathrm{F}_{2}(g) \rightarrow \mathrm{UF}_{6}(g) \end{array}$$.a. How many kilograms of HF are needed to completely react with \(5.00 \mathrm{kg}\) of \(\mathrm{UO}_{2} ?\) b. How much UF \(_{6}\) can be produced from \(850.0 \mathrm{g}\) of \(\mathrm{UO}_{2} ?\)

Step-by-Step Solution

Answer

Question: Calculate the mass (in kilograms) of HF needed to completely react with 5.00 kg of UO2. Also, calculate the mass of UF6 produced from 850.0 g of UO2. Answer: Approximately 1.482 kg of HF are needed to completely react with 5.00 kg of UO2. Approximately 1107.53 g of UF6 can be produced from 850.0 g of UO2.

Step 1: Find molar mass of each compound

First, we need to find the molar mass of UO2, HF, UF4, and UF6. To do this, sum the molar mass of each element in the compound and multiply it by its subscript. UO2: (1 x 238.03) + (2 x 16.00) = 270.03 g/mol HF: (1 x 1.01) + (1 x 19.00) = 20.01 g/mol UF4: (1 x 238.03) + (4 x 19.00) = 314.03 g/mol UF6: (1 x 238.03) + (6 x 19.00) = 352.03 g/mol

Step 2: a - Calculate moles of UO2

For part a, we start by calculating the moles of UO2 using the given mass (5.00 kg) and its molar mass. Convert mass from kg to g: 5.00 kg x (1000 g/kg) = 5000 g of UO2 moles of UO2 = mass / molar mass = 5000 g / 270.03 g/mol ≈ 18.518 mol of UO2

Step 3: a - Use mole ratio to calculate moles of HF needed

From the chemical equation UO2(g) + 4 HF(aq) → UF4(g) + 2 H2O(l), we see that 1 mole of UO2 reacts with 4 moles of HF. moles of HF needed = moles of UO2 * (4 mol HF / 1 mol UO2) ≈ 18.518 * 4 = 74.072 mol of HF

Step 4: a - Calculate mass of HF needed

Finally, we can find the mass of HF needed by multiplying the moles of HF by its molar mass. mass of HF = moles of HF * molar mass = 74.072 mol * 20.01 g/mol ≈ 1481.91 g Convert mass to kg: 1481.91 g x (1 kg/1000 g) ≈ 1.482 kg

Step 5: Answer (a):

Approximately 1.482 kg of HF are needed to completely react with 5.00 kg of UO2.

Step 2: b - Calculate moles of UO2

For part b, we start by calculating the moles of UO2 using the given mass (850.0 g) and its molar mass. moles of UO2 = mass / molar mass = 850.0 g / 270.03 g/mol ≈ 3.148 mol of UO2

Step 3: b - Use mole ratio to calculate moles of UF6 produced

From the chemical equation UF4(g) + F2(g) → UF6(g), we see that 1 mole of UF4 produces 1 mole of UF6. Since the first reaction results in 1:1 conversion between UO2 and UF4, the mole ratio of UO2 to UF6 is also 1:1. moles of UF6 produced = moles of UO2 ≈ 3.148 mol

Step 4: b - Calculate mass of UF6 produced

Finally, we can find the mass of UF6 produced by multiplying the moles of UF6 by its molar mass. mass of UF6 = moles of UF6 * molar mass = 3.148 mol * 352.03 g/mol ≈ 1107.53 g

Step 9: Answer (b):

Approximately 1107.53 g of UF6 can be produced from 850.0 g of UO2.

Key Concepts

Mole Concept

Understanding the mole concept is integral to mastering chemistry, especially when dealing with chemical reactions. A mole is a unit used to express the amount of a substance. It's similar to saying a 'dozen' when referring to 12 items. However, in chemistry, a mole refers to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities. This can include atoms, molecules, ions, or other particles.

Let's take the uranium enrichment problem as an example. In the solution, we converted the mass of \( \mathrm{UO}_{2} \) into moles to find out how much other substances, such as HF, would react with it. To do that, we first determined the molar mass of \( \mathrm{UO}_{2} \), which is the mass of one mole of that substance. Once we had the molar mass, we could convert the given kilograms into moles using the formula:

\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]
Thus, converting the mass to moles is the first crucial step in stoichiometry, which allows us to compute the amounts of reactants and products involved in chemical reactions.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It's derived from the Greek words 'stoicheion' (element) and 'metron' (measure), essentially meaning 'measuring elements'. Stoichiometry is based on the conservation of mass where the mass of the reactants equals the mass of the products.

In our uranium enrichment example, the stoichiometric calculations involve using the balanced chemical equations to find out how much \( \mathrm{HF} \) is needed to react with a given amount of \( \mathrm{UO}_{2} \) and how much \( \mathrm{UF}_{6} \) will be produced. The coefficients in the balanced equation represent the stoichiometric ratios, indicating the proportion in which the substances react or form.

For instance, the chemical equation indicates that 1 mole of \( \mathrm{UO}_{2} \) reacts with 4 moles of \( \mathrm{HF} \). We can express this as a mole ratio:

\[ \frac{4 \text{ mol HF}}{1 \text{ mol UO}_2} \]
This ratio becomes the bridge between moles of one substance to moles of another, enabling us to use it in stoichiometric calculations to determine the amounts needed or produced, as shown in the problem's solution.

Chemical Reaction Calculations

Chemical reaction calculations involve using the mole concept and stoichiometry to predict the outcomes of chemical reactions quantitatively. To carry out these calculations successfully, one must first write out a balanced chemical equation. The balanced equation is necessary because it shows the precise proportions in which reactants combine and products form, respecting the law of conservation of mass.

In the uranium enrichment process, the key calculation steps involve determining how many moles of substances are involved and then, based on the balanced equations, figuring out how many moles are needed or produced. After that, it's a matter of converting moles back into a measurable quantity, such as grams or kilograms. When performing the calculations, it's essential to be meticulous with units and conversion factors, as seen in the steps provided in the exercise.

To calculate the mass of a substance needed or produced, you can use the formula:

\[ \text{mass} = \text{moles} \times \text{molar mass} \]
By applying these calculated masses, one can assess the efficiency of a chemical process, estimate costs for industrial applications, and ensure the safety and effectiveness of the reactions. All these calculations derive from the foundational concepts of the mole and stoichiometry.