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Question #44665

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The Question

A $1.84 \text{ g}$ sample of pure Tungsten ( $\ce{W}$ , atomic mass $= 184 \text{ g/mol}$ ) is reacted with $1.064 \text{ g}$ of Fluorine gas ( $\ce{F2}$ , molar mass $= 38 \text{ g/mol}$ ) in a sealed, evacuated vessel at $600 \text{ K}$ . Tungsten exhibits variable oxidation states, primarily forming $\ce{WF5}$ and $\ce{WF6}$ depending on the stoichiometric ratio of the reactants. Assuming the reaction proceeds to completion such that no elemental tungsten or fluorine remains, what is the mole ratio of $\ce{WF6}$ to $\ce{WF5}$ in the final product mixture?

Options

A.
$3 : 2$
B.
$1 : 0$
C.
$0 : 1$
D.
$2 : 3$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Limiting Reagent Stoichiometry|D-block Variable Oxidation States|Stability of High O.S. Fluorides/Oxides

Why This Question Is Tricky

Common Mistakes

This question tests the 'Limiting Reagent Trap' in the context of d-block variable oxidation states.

Students typically assume a single reaction (e.g., $\ce{W} + 3\ce{F2} \rightarrow \ce{WF6}$ ) and calculate that $\ce{F2}$ is the limiting reagent, incorrectly concluding that only $\ce{WF6}$ forms (Option B) and some Tungsten is left unreacted.

The 'Hidden Variable' is the competitive thermodynamic stability of lower oxidation states (like $\ce{WF5}$ ).

When the halogen is not in sufficient excess to push the metal entirely to its highest oxidation state, a mixture of oxidation states forms to fully consume the reactants, invalidating standard single-product limiting reagent assumptions.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Some Basic Concepts of Chemistry - Class 11 Part 1
Concept	Limiting Reagent Stoichiometry
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free