1. Write the Balanced Chemical Equation

The reaction between xenon (Xe) and fluorine (F₂) to form xenon tetrafluoride (XeF₄) is:

Xe + 2F₂ → XeF₄

2. Determine the Limiting Reactant

* Calculate moles of Xe:

* Molar mass of Xe = 131.29 g/mol

* Moles of Xe = 130 g / 131.29 g/mol = 0.99 mol

* Calculate moles of F₂:

* Molar mass of F₂ = 38.00 g/mol

* Moles of F₂ = 100 g / 38.00 g/mol = 2.63 mol

* Compare the mole ratio: The balanced equation shows that 1 mole of Xe reacts with 2 moles of F₂. Since we have 0.99 moles of Xe and 2.63 moles of F₂, Xe is the limiting reactant because we don't have enough F₂ to react with all of the Xe.

3. Calculate the Theoretical Yield of XeF₄

* The balanced equation shows that 1 mole of Xe produces 1 mole of XeF₄.

* Since we have 0.99 moles of Xe, we can theoretically produce 0.99 moles of XeF₄.

4. Convert Moles of XeF₄ to Grams

* Molar mass of XeF₄ = 207.29 g/mol

* Mass of XeF₄ = 0.99 mol * 207.29 g/mol = 205.26 g

Therefore, the theoretical mass of xenon tetrafluoride that should form is 205.26 g.