Concept:In reactions involving multiple reactants, the limiting reagent is the one that is entirely consumed first, strictly dictating the maximum amount of product that can be formed.
Formula:$$ \text{Limiting Reagent} = \text{Smallest value of } \frac{\text{Moles supplied}}{\text{Stoichiometric Coefficient}} $$
Solution:- Calculate moles supplied:
- \( n(\text{XeF}_6) = \frac{122.6}{245.3} = 0.5 \text{ moles} \).
- \( n(\text{SiO}_2) = \frac{60}{60} = 1.0 \text{ moles} \).
- Divide by coefficients to find Limiting Reagent:
- For \( \text{XeF}_6 \): \( 0.5 / 2 = 0.25 \). (This is the smaller value, making it the Limiting Reagent).
- For \( \text{SiO}_2 \): \( 1.0 / 1 = 1.0 \).
- Since \( \text{XeF}_6 \) limits the reaction, use it to calculate \( \text{SiF}_4 \).
- From the equation: 2 moles \( \text{XeF}_6 \) produce 1 mole \( \text{SiF}_4 \). Therefore, 0.5 moles of \( \text{XeF}_6 \) produce 0.25 moles of \( \text{SiF}_4 \).
- Mass of \( \text{SiF}_4 \) produced = \( 0.25 \text{ moles} \times 104 \text{ g/mol} = 26 \text{ g} \).
Why other options are incorrect:- Option B: Incorrectly identifies \( \text{SiO}_2 \) as the limiting reagent.
- Option C & D: Yield exactly double the mass (52g) because they assume 0.5 moles of \( \text{SiF}_4 \) was generated, completely ignoring the 2:1 stoichiometric ratio.
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