Question

The mass of oxygen that would be required to produce enough co which completely reduces 1.6 kg fe2o3

Answer 1

A balance chemical equation of Fe₂O₃ and CO :
Fe₂O₃ + 3CO ⇔2Fe + 3CO₂
Here it is clear that, 1 mole of Fe₂O₃ reacts with 3moles of CO.
or, (56× 2 + 48)=160g of Fe₂O₃ reacts with 3 × 28 = 84g of CO.
or, 1.6kg of Fe₂O₃ reacts with 84/160 × 1.6 = 840g of CO

Now, again see formation of CO by chemical balance equation :
C + 1/2O₂ ⇔ CO
Here it is clear that , 1 mole of CO is produced by 0.5 mole of O₂.
or, 28g of CO is produced by 16g of O₂ .
or, 840 g of CO is produced by 16/28 × 840 = 480g of O₂

Hence, weight of oxygen = 480g

Answer 2

Hello Dear.

Here is the answer---


 Reaction of Fe₂O₃ with CO is →

Fe₂O₃ + 3CO --------→ 2Fe +3CO₂

Molar Mass of Fe₂O₃ = 160 g/mole.
Molar mass of CO = 28 g/mole.

Now,
From the Reaction,
∵ 160 g of Fe₂O₃ reacts with 3 × 28 g of CO.
∴ 1 -----------------------------84/160 g of CO.
∴ 1.6 kg-----------------------(0.525) × 1600 g of CO.
∴ Mass of CO which Reacts with 1.6 kg of Fe₂O₃ = 840 g.


Now, Reaction to show the Formation of Carbon Monoxide.

 2C + O₂ -------→ 2CO

From the Reaction,

∵ 2 mole of CO is produced by 1 mole of O₂.
∴ 2 × 28 g of CO is produced by the 32 g of O₂.
∴ 1 g of CO is produced by the 32/56 g of O₂.
∴ 840 g of CO is Produced by the (4/7) × 840 g of O₂
     = 4 × 120 g of O₂.
     = 480 grams of O₂

Thus, Mass of Oxygen that would be required to Produce 840 g CO which completely reduces 1600 g of Fe₂O₃ is 480 g.


Hope it helps.