Problem 39 - Chemistry: An Atoms-Focused Approach

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 UO, with HF at high temperatures to form UF \(_{4},\) which is then converted into UF \(_{6}\) in another high-temperature reaction with fluorine gas: $$(1)\mathrm{UO}_{2}(g)+4 \mathrm{HF}(a q) \rightarrow \mathrm{UF}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$$ $$(2)\mathrm{UF}_{4}(g)+\mathrm{F}_{2}(g) \rightarrow \mathrm{UF}_{6}(g)$$ a. How many kilograms of HF are needed to completely react with \(5.00 \mathrm{kg} \mathrm{UO}_{2} ?\) b. How much UF \(_{6}\) can be produced from \(850.0 \mathrm{g} \mathrm{UO}_{2} ?\)

Step-by-Step Solution

Answer

Answer: To completely react with 5.00 kg of UO2, 1.68067 kg of HF is needed. From 850.0 g of UO2, 1257.14 g of UF6 can be produced.

Step 1: Calculate the moles of UO2 given in the problems#a. 5.00 kg UO2 b. 850.0 g UO2

First, we need to convert the given masses of UO2 into moles. To do this, we'll use the molar mass of UO2: Molar mass of UO2 = 92.0 g/mol U + (2 * 16.0 g/mol O) = 238.0 g/mol Now we can convert: a. Moles of UO2 = (5.00 kg UO2 * 1000 g/kg) / 238.0 g/mol = 21.0084 moles UO2 b. Moles of UO2 = 850.0 g UO2 / 238.0 g/mol = 3.57143 moles UO2

Step 2: Use stoichiometry to find moles of other compounds involved in the reactions

From the balanced chemical equations, we see that 1 mole of UO2 requires 4 moles of HF for the first reaction and 1 mole of UF4 requires 1 mole of F2 for the second reaction. a. Moles of HF needed = 21.0084 moles UO2 * 4 moles HF / 1 mole UO2 = 84.0336 moles HF b. First, we need to find the moles of UF4 produced from UO2: Moles of UF4 = 3.57143 moles UO2 * 1 mole UF4 / 1 mole UO2 = 3.57143 moles UF4 Now, we can find the moles of UF6 that can be produced from UF4: Moles of UF6 = 3.57143 moles UF4 * 1 mole UF6 / 1 mole UF4 = 3.57143 moles UF6

Step 3: Convert moles to mass using molar masses of the compounds

Now that we have the moles of the compounds, we can convert them back to mass using their molar masses: Molar mass of HF = 1.0 g/mol H + 19.0 g/mol F = 20.0 g/mol Molar mass of UF4 = 238.0 g/mol U + (4 * 19.0 g/mol F) = 314.0 g/mol Molar mass of UF6 = 238.0 g/mol U + (6 * 19.0 g/mol F) = 352.0 g/mol a. Mass of HF = 84.0336 moles HF * 20.0 g/mol = 1680.67 g = 1.68067 kg b. Mass of UF6 = 3.57143 moles UF6 * 352.0 g/mol = 1257.14 g

Step 4: Write the final answers

Taking the results from the previous steps, we can present the final answers for each part: a. To completely react with 5.00 kg of UO2, 1.68067 kg of HF is needed. b. From 850.0 g of UO2, 1257.14 g of UF6 can be produced.

Key Concepts

Stoichiometry

Stoichiometry is the tool we use to understand the quantitative relationships in chemical reactions. It tells us how much of each reactant is needed and how much product is formed in a reaction. To solve stoichiometry problems, the balanced chemical equation is crucial. It shows the ratio of reactants to products. Take the first reaction in uranium enrichment:

• \( \text{UO}_2 + 4 \text{HF} \rightarrow \text{UF}_4 + 2 \text{H}_2\text{O} \)

This equation tells us that 1 mole of \( \text{UO}_2 \) reacts with 4 moles of \( \text{HF} \) to produce \( \text{UF}_4 \) and water. By using stoichiometric ratios, we can convert between moles of different substances based on the coefficients from the balanced equation. This conversion is essential to predict the amounts of reactants consumed and products formed in a reaction, allowing us to solve various quantitative problems in chemistry.

Chemical Reactions

Chemical reactions are processes where substances, called reactants, transform into new substances, called products. In uranium enrichment, several chemical reactions occur. The key reactions involve changing uranium to forms that can be further processed or utilized:

• The first reaction involves \( \text{UO}_2 \) reacting with \( \text{HF} \), producing \( \text{UF}_4 \), a solid, and \( \text{H}_2\text{O} \), a liquid.
• The second reaction is the conversion of \( \text{UF}_4 \) into \( \text{UF}_6 \) using fluorine gas (\( \text{F}_2 \)).

In such reactions, it’s crucial to balance the chemical equations to adhere to the law of conservation of mass. Proper balancing ensures that the number of atoms for each element is the same on both the reactant and product sides of the reaction. This balance helps in calculating precise reactant amounts needed and product quantities formed.

Nuclear Fuel

Uranium is a critical element used as nuclear fuel, particularly in the form of \(^{235}\text{U} \), which is an isotope of uranium capable of sustaining nuclear fission. Its enrichment process involves increasing the percentage of \(^{235}\text{U} \) relative to \(^{238}\text{U} \), which naturally occurs more abundantly. The chemical reactions involved in enrichment change uranium into forms that make this separation or concentration feasible:

• One step is the conversion of uranium dioxide \(\text{UO}_2\) to uranium tetrafluoride \(\text{UF}_4\).
• This is then further transformed into uranium hexafluoride \(\text{UF}_6\), a gas suitable for processes like gaseous diffusion or centrifugation.

These conversions and the resulting chemical form (\( \text{UF}_6 \)) are vital as they allow for the efficient and precise handling and enrichment of uranium isotopes for use in nuclear reactors, ensuring a concentrated source of \(^{235}\text{U} \) in a manageable form for fuel production.

Molar Mass Calculations

Molar mass calculations are fundamental to converting between mass and moles in chemistry, enabling the quantitative analysis of chemical reactions. To perform these calculations, one needs to sum up the atomic masses of each element present in a compound:

• For \( \text{UO}_2 \), the molar mass is calculated by summing the mass of uranium \((92.0 \text{ g/mol})\) and twice the mass of oxygen \((2 \times 16.0 \text{ g/mol})\), resulting in \(238.0 \text{ g/mol}.\)
• Similarly, for \( \text{UF}_6 \), we combine the mass of uranium \((238.0 \text{ g/mol})\) with six times the mass of fluorine \((6 \times 19.0 \text{ g/mol})\) to get \(352.0 \text{ g/mol}.\)

With these molar masses, we can convert given masses (in grams or kilograms) of substances into moles, essential for stoichiometric calculations. For example, converting \(5.00 \text{ kg} \text{ of } \text{UO}_2\) into moles allows us to figure out how many moles of \( \text{HF} \) are needed or how much \( \text{UF}_6 \) can be produced.