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)
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To find the number of moles of CsXeF7 that can be produced from the reaction, we need to determine the limiting reactant.
1. Convert the given amounts of cesium fluoride (CsF) and xenon hexafluoride (XeF6) to moles.
- Moles of CsF = 13.0 mol
- Moles of XeF6 = 14.0 mol
2. Write and balance the equation for the reaction:
CsF(s) + XeF6(s) → CsXeF7(s)
3. Determine the stoichiometry of the reaction by comparing the coefficients from the balanced equation:
- The coefficient of CsF is 1.
- The coefficient of XeF6 is 1.
- The coefficient of CsXeF7 is also 1.
4. Use the stoichiometry to determine the limiting reactant:
- The reactant with the smaller stoichiometric coefficient is the limiting reactant.
- In this case, both reactants have a stoichiometric coefficient of 1, so we need to compare their initial moles.
Moles of CsF = 13.0 mol
Moles of XeF6 = 14.0 mol
Since both reactants have a stoichiometric coefficient of 1, we can directly compare their moles. The reactant with fewer moles is the limiting reactant.
The limiting reactant is CsF with 13.0 mol.
5. Use the stoichiometry to calculate the maximum moles of CsXeF7 that can be produced.
- Since the stoichiometric coefficient of CsXeF7 is 1, the maximum moles of CsXeF7 formed will be the same as the moles of the limiting reactant, which is 13.0 mol.
Therefore, 13.0 moles of CsXeF7 can be produced from the reaction of 13.0 mol cesium fluoride with 14.0 mol xenon hexafluoride.