Question

How many moles of \(\mathrm{UF}_{6}\) would have to be decomposed to provide enough fluorine to prepare \(1.25 \mathrm{~mol}\) of \(\mathrm{CF}_{4} ?\) (Assume sufficient carbon is available.)

Short answer

0.83 moles of \(\mathrm{UF}_{6}\) is needed to decompose to provide enough fluorine to prepare 1.25 mol of \(\mathrm{CF}_{4}\).

Step-by-step solution

Write the balanced chemical equation

To find out how many moles of \(\mathrm{UF}_{6}\) are needed, start by writing a balanced chemical equation for the reaction where \(\mathrm{UF}_{6}\) decomposes to release \(F_2\) and then the \(F_2\) reacts with carbon (C) to form \(\mathrm{CF}_{4}\). The equations can be represented as: \[ \mathrm{UF}_{6} \rightarrow \mathrm{U} + 3\mathrm{F}_{2} \] \[ \mathrm{C} + 2\mathrm{F}_{2} \rightarrow \mathrm{CF}_{4} \]

Determine moles of \(F_2\) needed

From the second reaction, 1 mole of \(\mathrm{CF}_{4}\) requires 2 moles of \(F_2\). So, to prepare 1.25 moles of \(\mathrm{CF}_{4}\), the moles of \(F_2\) required would be: \[ \mathrm{Moles\ of\ F}_{2} = 1.25 \,\mathrm{mol} \times \dfrac{2 \, \mathrm{mol\ F}_{2}}{1 \, \mathrm{mol\ CF}_{4}} = 2.50 \,\mathrm{mol} \]

Calculate moles of \(\mathrm{UF}_{6}\) needed

From the first reaction, 1 mole of \(\mathrm{UF}_{6}\) produces 3 moles of \(F_2\). To find out how many moles of \(\mathrm{UF}_{6}\) is needed to obtain 2.50 moles of \(F_2\), use the following ratio: \[ \mathrm{Moles\ of\ UF}_{6} = 2.50 \,\mathrm{mol\ F}_{2} \times \dfrac{1 \, \mathrm{mol\ UF}_{6}}{3 \, \mathrm{mol\ F}_{2}} = 0.83 \,\mathrm{mol} \]

Key concepts

Chemical Equation Balancing

Understanding how to balance a chemical equation is fundamental to solving stoichiometry problems. It's much like ensuring both sides of a scale are equal. In the exercise, balanced equations are critical for figuring out how much of each substance is involved in the chemical reactions. The equation \[ \text{UF}_{6} \rightarrow \text{U} + 3\text{F}_{2} \] represents the decomposition of uranium hexafluoride (\text{UF}_{6}), where each molecule breaks down into one atom of uranium (U) and three molecules of fluorine gas (\text{F}_{2}). Note that '3' before \text{F}_{2} indicates three molecules of fluorine are produced for every one molecule of uranium hexafluoride that decomposes.

Likewise, the formation of carbon tetrafluoride (\text{CF}_{4}) from carbon (C) and fluorine gas is depicted by\[ \text{C} + 2\text{F}_{2} \rightarrow \text{CF}_{4} \]In this reaction, one atom of carbon reacts with two molecules of fluorine gas to create one molecule of carbon tetrafluoride. Here again, balancing ensures that the atoms of each element are conserved according to the principle of conservation of mass. Mastering the skill of balancing equations is not just an academic exercise; it lays the groundwork for predictive control of chemical reactions in various applications from industrial synthesis to pharmaceuticals.

Stoichiometric Calculations

Stoichiometric calculations tell us how reactants relate to products and to each other. It's a bit like a recipe where ingredients must be mixed in the right proportions to get the desired outcome. When we know the ratio of reactants to products, as given by the balanced chemical equations, we can calculate the actual amounts needed or produced. In our example, we calculated the moles of fluorine gas (\text{F}_{2}) needed to produce 1.25 mol of carbon tetrafluoride (\text{CF}_{4}) using the stoichiometric ratio from the balanced equation.

Here's how it was done:\[ \text{Moles of F}_{2} = 1.25 \text{ mol} \times \frac{2 \text{ mol F}_{2}}{1 \text{ mol CF}_{4}} = 2.50 \text{ mol} \]This calculation is crucial because it translates the proportion from the equation into an actual quantity we can measure and use. Following this, the amount of \text{UF}_{6} needed was found through a similar process, taking into account the relationship between \text{UF}_{6} decomposing to form \text{F}_{2}. These step-by-step connections allow scientists and engineers to predict the outcomes of reactions and design processes accordingly.

Mole Concept

The mole concept is a bridge that connects the world of atoms and molecules with the world we can measure, allowing chemists to 'count' by weighing. One mole is Avogadro's number (\(6.022 \times 10^{23}\)) of particles, be it atoms, molecules, or ions. Understanding the mole is like having a universal translator for chemistry. When we speak of 1.25 mol of \text{CF}_{4}, we know it contains Avogadro's number of molecules multiplied by 1.25. This concept lets us convert between masses and number of particles through molar mass – the mass of one mole of a substance.

In practical terms, the mole allows us to measure out exact amounts of substances for reactions. By using the mole concept, the solution to our exercise calculated\[ \text{Moles of UF}_{6} = 2.50 \text{ mol F}_{2} \times \frac{1 \text{ mol UF}_{6}}{3 \text{ mol F}_{2}} = 0.83 \text{ mol} \]This shows how the theoretical understanding of the mole concept is applied to calculate the precise amount of \text{UF}_{6} needed to produce the desired quantity of fluorine for \text{CF}_{4} synthesis. The mole is essential not just in educational exercises, but also in real-world chemical production and laboratory experiments.