# FeO (s) + CO (g) = Fe (s) + CO2 (g), k = 0. 435 at 1173E; at equilibrium, what will be the number of moles of CO gas required to reduce one mole of FeO at 1173 K ? with Answer - Chemical Engineering Mcqs | TestPrep

> Solved MCQ: FeO (s) + CO (g) = Fe (s) + CO2 (g), k = 0. 435 at 1173E; at equilibrium, what will be the number of moles of CO gas required to reduce one mole of FeO at 1173 K ? Correct Answer: D. 3.3. Verified answer explanation for FPSC, PPSC, MDCAT, NTS, and CSS competitive exam preparation.

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# FeO (s) + CO (g) = Fe (s) + CO2 (g), k = 0. 435 at 1173E; at equilibrium, what will be the number of moles of CO gas required to reduce one mole of FeO at 1173 K ?

Direct Answer & Quick Reference

The correct option is **D. 3.3**.

A. 1.0

B. 1.3

C. 2.3

D. 3.3 **(Correct Answer)**

✨ Explain with AI

Verified Key: D. 3.3
Solved by 3,943+ students

Verified against official examination board answer keys and standard curriculum textbooks.

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*Source: https://test-prep.blog/mcq/feo-s-co-g-fe-s-co2-g-k-0-435-at-1173e-at-equilibrium-what-will-be-the-number-of-e0eb19c6-c16a-43f8-8a82-285522fb9666*
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