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Question #44678
The Question
A sample of pure iron pyrites (\ce{FeS2}) is roasted in excess oxygen. The entire \ce{SO2} gas produced is collected and dissolved in water. To this solution, of acidified \ce{K2Cr2O7} is added, which completely reacts with a portion of the \ce{SO2}. Subsequently, of \ce{H2S} gas is bubbled through the remaining solution, resulting in the precipitation of elemental sulfur (\ce{S}). Assuming no other side reactions, determine the mass of elemental sulfur precipitated and the mass of the excess reagent remaining in the final step.\n\n(Molar masses in : \ce{FeS2} = 120, \ce{SO2} = 64, \ce{H2S} = 34, \ce{S} = 32)
Options
- A.
of \ce{S} ; of \ce{SO2} remains
- B.
of \ce{S} ; of \ce{H2S} remains
- C.
of \ce{S} ; of \ce{SO2} remains
- D.
of \ce{S} ; of \ce{H2S} remains
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Mechanism_Trap
Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.
Concepts Behind This Question
Equivalent Concept in Redox|Limiting Reagent Stoichiometry
Why This Question Is Tricky
Common Mistakes
This problem tests the 'Variable n-factor' trap in consecutive redox reactions.
Intermediate \ce{SO2} acts as a reductant in Step 1 (n-factor = 2) and an oxidant in Step 2 (n-factor = 4).
Students who mechanically apply the equivalent concept often carry over the n-factor of 2 into the second step.
This false assumption calculates equivalents of \ce{SO2}, making it appear as the limiting reagent against equivalents of \ce{H2S}.
This flips the excess reagent from \ce{SO2} to \ce{H2S} and yields an incorrect mass of precipitated sulfur (Option B).
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Some Basic Concepts of Chemistry - Class 11 Part 1 |
| Concept | Equivalent Concept in Redox |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |