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

FO (s) + CO (g) = Fe (s) + CO (g), = 0. 435 at 1173E; At equilibrium,

1.	FO (s) + CO (g) = Fe (s) + CO (g), = 0. 435 at 1173E; At equilibrium, what will be the number of moles of CO gas required to reduce one mole of FO at 1173 K?
A.	1.0
B.	1.3
C.	2.3
D.	3.3
Answer» E.

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

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