Blast furnaces extra pure iron from the Iron(IIl)oxide in iron ore in a two step sequence. In the first step, carbon and oxygen react to form carbon monoxide:
2C(s)+O2(g) arrow 2CO(g)
In the second step, iron(lll) oxide and carbon monoxide react to form Iron and carbon dioxide:
Fe203(s) + 3CO(g) arrow 2Fe(s)+ 3CO2(g)
Suppose the yield of the first step is 71.% and the yield of the second step is 72.%. Calculate the mass of oxygen required to make 7.0 kg of iron.
Be sure your answer has a unit symbol, if needed, and is rounded to 2 significant digits.

5.9 kg  

Explanation:

We must work backwards from the second step to work out the mass of oxygen.

1. Second step

Mᵣ:                                     55.84

            Fe₂O₃ + 3CO  ⟶  2Fe  +  3CO₂

m/kg:                                    7.0

(a) Moles of Fe

(b) Moles of CO

However, this is the theoretical yield.

The actual yield is 72. %.

We need more CO and Fe₂O₃ to get the theoretical yield of Fe.

(c) Percent yield

We must use 261 mol of CO to get 7.0 kg of Fe.

2. First step

Mᵣ:                32.00

            2C   +  O₂   ⟶  2CO

n/mol:                             261

(a) Moles of O₂

(b) Mass of O₂

However, this is the theoretical yield.

The actual yield is 71. %.

We need more C and O₂ to get the theoretical yield of CO.

(c) Percent yield

We need 5.9 kg of O₂ to produce 7.0 kg of Fe.

there all right good

explanation:

the answer is definitely c.