Question

crystals of xenon tetrafluoride
if 0.880 grams of $\ce{f_{2}}$ are produced by the above reaction, what mass of $\ce{xef_{4}}$ reacted?
report your answer to two decimal places.
show molar masses of reactants and products
\boxed{} grams $\ce{xef_{4}}$

Explanation:

Step1: Determine the reaction (assumed decomposition of XeF₄: \( \text{XeF}_4

ightarrow \text{Xe} + 2\text{F}_2 \))
Molar mass of \( \text{F}_2 \): \( 2 \times 19.00 = 38.00 \, \text{g/mol} \)
Molar mass of \( \text{XeF}_4 \): \( 131.29 + 4 \times 19.00 = 207.29 \, \text{g/mol} \)

Step2: Moles of \( \text{F}_2 \) produced

\( \text{Moles of } \text{F}_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.880 \, \text{g}}{38.00 \, \text{g/mol}} \approx 0.02316 \, \text{mol} \)

Step3: Moles of \( \text{XeF}_4 \) reacted (from stoichiometry: 1 mol \( \text{XeF}_4 \) produces 2 mol \( \text{F}_2 \))

\( \text{Moles of } \text{XeF}_4 = \frac{0.02316 \, \text{mol} \, \text{F}_2}{2} = 0.01158 \, \text{mol} \)

Step4: Mass of \( \text{XeF}_4 \) reacted

\( \text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.01158 \, \text{mol} \times 207.29 \, \text{g/mol} \approx 2.39 \, \text{g} \)

Answer:

2.39