One of the few xenon compounds that form is cesium xenon heptafluoride (CsXeF7). How many moles of CsXeF7 can be produced from the reaction of 13.0 mol cesium fluoride with 14.0 mol xenon hexafluoride?
CsF(s) + XeF6(s) --> CsXeF7(s)
To determine the number of moles of CsXeF7 that can be produced from the reaction of 13.0 mol cesium fluoride (CsF) and 14.0 mol xenon hexafluoride (XeF6), we need to compare the mole ratio of the reactants to the product.
From the balanced equation:
1 mole of CsF reacts with 1 mole of XeF6 to produce 1 mole of CsXeF7.
Therefore, the limiting reactant will determine the number of moles of CsXeF7 that can be produced. The reactant with the smaller mole ratio compared to the stoichiometric ratio will be the limiting reactant.
The mole ratio of CsF to CsXeF7 is 1:1, so 13.0 mol of CsF will produce 13.0 mol of CsXeF7.
The mole ratio of XeF6 to CsXeF7 is also 1:1, so 14.0 mol of XeF6 will produce 14.0 mol of CsXeF7.
Since the mole ratio of XeF6 to CsXeF7 is greater than that of CsF to CsXeF7, CsF is the limiting reactant.
Therefore, 13.0 mol of CsXeF7 can be produced from the reaction.
To determine the number of moles of CsXeF7 that can be produced from the given reaction, we need to use the concept of stoichiometry. Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction.
1. Write the balanced equation:
CsF(s) + XeF6(s) → CsXeF7(s)
2. Determine the mole ratio between CsXeF7 and the reactants based on the balanced equation:
From the balanced equation, we can see that the mole ratio between CsXeF7 and CsF is 1:1, and the mole ratio between CsXeF7 and XeF6 is also 1:1.
3. Determine the limiting and excess reactants:
To determine the limiting reactant, compare the moles of each reactant given in the problem. The reactant that produces the least amount of product is the limiting reactant.
- Moles of CsF: 13.0 mol
- Moles of XeF6: 14.0 mol
Since the mole ratio between CsF and CsXeF7 is 1:1, and 13.0 mol of CsF is given, it means that 13.0 mol of CsXeF7 can be produced.
However, since the mole ratio between XeF6 and CsXeF7 is also 1:1, and 14.0 mol of XeF6 is given, it means that 14.0 mol of CsXeF7 can be produced.
Therefore, the limiting reactant is CsF because it produces a lesser amount of product.
4. Calculate the moles of CsXeF7 produced:
Since the limiting reactant is CsF, we know that 13.0 mol of CsXeF7 can be produced.
Thus, 13.0 moles of CsXeF7 can be produced from the reaction of 13.0 mol cesium fluoride with 14.0 mol xenon hexafluoride.