Question

One of the few xenon compounds that form is cesium xenon heptafluoride (CsXeF \(_{7} ) .\) How many moles of CsXeF \(_{7}\) can be produced from the reaction of 12.5 \(\mathrm{mol}\) of cesium fluoride with 10.0 \(\mathrm{mol}\) of xenon hexafluoride? $$\mathrm{CsF}(\mathrm{s})+\mathrm{XeF}_{6}(\mathrm{s}) \rightarrow \mathrm{Cs} \mathrm{XeF}_{7}(\mathrm{s})$$

Short answer

The maximum number of moles of cesium xenon heptafluoride (CsXeF7) that can be produced from the reaction of 12.5 mol of cesium fluoride with 10.0 mol of xenon hexafluoride is 10.0 moles.

Step-by-step solution

Write the balanced chemical equation

First, write down the balanced chemical equation for the reaction: \[ \mathrm{CsF}(\mathrm{s})+\mathrm{XeF}_{6}(\mathrm{s}) \rightarrow \mathrm{Cs}\mathrm{XeF}_{7}(\mathrm{s})\] We can see that the ratio between the reactants and the product is 1:1 in this balanced chemical equation.

Calculate the moles of each reactant needed for the reaction to proceed

According to the balanced chemical equation, 1 mole of CsF is needed to react with 1 mole of XeF6 to produce 1 mole of CsXeF7. We have been given the moles of both reactants: Moles of CsF = 12.5 mol Moles of XeF6 = 10.0 mol

Determine the limiting reactant

To determine the limiting reactant, compare the ratio of moles of each reactant available with the ratio required according to the balanced chemical equation. Moles of CsF required for the reaction to proceed with the given moles of XeF6 is: Moles CsF = Moles XeF6 Moles CsF = 10.0 mol Since we have more moles of CsF than needed (12.5 > 10.0), XeF6 is the limiting reactant in this reaction.

Calculate the moles of CsXeF7 formed

Using the limiting reactant, calculate the moles of CsXeF7 that can be formed according to the balanced chemical equation: Moles of CsXeF7 = Moles of XeF6 Moles of CsXeF7 = 10.0 mol

Present the final answer

The maximum number of moles of cesium xenon heptafluoride (CsXeF7) that can be produced from the reaction of 12.5 mol of cesium fluoride with 10.0 mol of xenon hexafluoride is 10.0 moles.

Key concepts

Limiting Reactant

Understanding the concept of a limiting reactant is essential for predicting the amount of product that can be formed in a chemical reaction. A limiting reactant is the substance that is totally consumed when the chemical reaction is complete. Once this reactant is used up, the reaction stops, and no more product can be formed, regardless of the amounts of other reactants available.

In our exercise, the reactants are cesium fluoride (CsF) and xenon hexafluoride (XeF6). To find out which is the limiting reactant, we compare the amount of moles of each reactant with the stoichiometric ratio from the balanced chemical equation. The stoichiometric ratio determines how much of each reactant is needed to produce the product. If we have 12.5 moles of CsF and 10.0 moles of XeF6, but the reaction requires 1 mole of CsF per mole of XeF6, then XeF6 becomes the limiting reactant because we only have enough XeF6 to react with 10.0 moles of CsF. Any CsF in excess does not affect the amount of product formed because it cannot react without sufficient XeF6.

To reinforce the concept, consider a simple analogy: If we are making sandwiches and we have 10 slices of bread (XeF6) and more than enough ham slices (CsF), the number of sandwiches we can make is limited by the amount of bread available. Once the bread is gone, no more sandwiches can be made, even if ham is left over. This helps to visualize why the total product (sandwiches or CsXeF7) is directly limited by the reactant that runs out first, in this case, XeF6.

Chemical Equation Balancing

Balancing chemical equations is a foundational skill in chemistry that ensures the law of conservation of mass is upheld. According to this law, mass is neither created nor destroyed during a chemical reaction. Therefore, the number of atoms for each element must be the same on both sides of the equation.

For the reaction to produce cesium xenon heptafluoride (CsXeF7), we have to ensure that the number of Cs, Xe, and F atoms in the reactants equals the number of each in the product. In our given equation, \[ \mathrm{CsF}(\mathrm{s}) + \mathrm{XeF}_{6}(\mathrm{s}) \rightarrow \mathrm{Cs}\mathrm{XeF}_{7}(\mathrm{s}) \],it is balanced with a 1:1 stoichiometric ratio, meaning one mole of CsF reacts with one mole of XeF6 to yield one mole of CsXeF7.

To build competence in balancing equations, a helpful practice is to systematically adjust coefficients (numbers placed before compounds) to balance the elements one at a time. Starting with the most complex molecule often makes the process smoother. Remember, we cannot change the subscripts in the chemical formulas as that would alter the substances themselves; we can only adjust the overall proportions (coefficients) of the reactants and products.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure and observe. One mole is defined as the amount of substance containing Avogadro's number of entities (atoms, molecules, ions, etc.), which is approximately \(6.022 \times 10^{23}\). Understanding this allows chemists to count particles by weighing them since knowing the number of moles gives information about the number of particles.

In our exercise, moles translate the weight of each compound into a count of molecules that form the reactant pair CsF and XeF6. When we say that there are 12.5 moles of CsF and 10.0 moles of XeF6, we're essentially dealing with huge numbers of units capable of reacting with each other. Using the mole concept, chemists can predict how much product results from known quantities of reactants and vice versa.

A practical tip to enhance understanding: visualize the mole concept using discrete items like marbles. If 1 mole equals a certain number of marbles, then discussing moles of a substance is just like counting out that specific number of marbles to use in a game or a pattern, tying the abstract number back to a tangible quantity.