Question #56364

Pure iron can be extracted from iron ore, Fe203, from the reduction of carbon monoxide as shown by the following sequence of steps.
Step 1 c + O2 --> CO2
Step 2 CO2 + c --> 2CO
Step 3 Fe2O3 + 3CO --> 2Fe + 3CO2
In a single run, what is the maximum mass of iron that can be extracted from 60.0 g of carbon?

Expert's answer

Let's combine first two reactions in order to see how much of CO can be formed from the carbon. So summarizing the reactions, and excluding the CO2 (because its presented on both sides) we obtain the following chemical reaction.

C + O2 + CO2 + C = CO2 + 2CO

C + O2 + CO2 + C = CO2 + 2CO

2C + O2 = 2CO

So the total amount of the produced CO is equal to the amount of the initial carbon. We should calculate the amount of CO.

n(CO) = n(C) = m(C)/M(C) = 60/12 = 5 mol

From the reduction reaction we see that the amount of Fe is 1.5 less than the amount of CO. Using formulas, we must calculate the amount of iron, and then its mass.

Fe2O3 + 3CO = 2Fe + 3CO2

n(Fe) = [2*n(CO)]/3 =2*5/3=3.33 mol

m(Fe) = n(Fe)*M(Fe) = 3.33*56 =186.67 g

C + O2 + CO2 + C = CO2 + 2CO

C + O2 + CO2 + C = CO2 + 2CO

2C + O2 = 2CO

So the total amount of the produced CO is equal to the amount of the initial carbon. We should calculate the amount of CO.

n(CO) = n(C) = m(C)/M(C) = 60/12 = 5 mol

From the reduction reaction we see that the amount of Fe is 1.5 less than the amount of CO. Using formulas, we must calculate the amount of iron, and then its mass.

Fe2O3 + 3CO = 2Fe + 3CO2

n(Fe) = [2*n(CO)]/3 =2*5/3=3.33 mol

m(Fe) = n(Fe)*M(Fe) = 3.33*56 =186.67 g

## Comments

## Leave a comment