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A blast furnace makes pig iron containing $3.6\% \; \ce{C}$, $1.4\% \; \ce{Si}$, and $95\% \; \ce{Fe}$. The ore is $80\% \; \ce{Fe2O3}$, $12\% \; \ce{SiO2}$, and $\ce{Al2O3}$. The blast furnace gas contains $28\% \; \ce{CO}$ and $12\% \; \ce{CO2}$. Assume that there is no iron loss through the slag. The atomic mass of $\ce{Fe}$, $\ce{Si}$, and $\ce{Ca}$ are $56$, $28$, and $\pu{40 g/mol}$, respectively.

  1. What is the weight of the ore used per ton of pig iron?
  2. What is the volume of the blast furnace gas produced per ton of pig iron?

The first question involved finding the weight of ore used (Exact quote: "The weight of the ore used per ton of pig iron is _"). I was able to do that using algebraic method to generate equations. The method that I applied was very much similar to this. However, a follow up question is asked about the volume of the blast furnace gas (Exact quote: "The volume of the blast furnace gas produced per ton of pig iron is___"). I do not know how to do that. All that I would prefer is to use a similar approach that has been used.

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As there is no iron loss through the slag, all your calculations must be made with respect to the iron atom, which is kept constant through all the process. So first, calculate the mass of pure iron in 1 ton pig iron. Then calculate the amount of substance of iron element in this mass. Then divide by two to obtain the amount of substance of iron oxide in the ore. Then calculate the mass of iron oxide in these amounts of iron oxides. Then divide by $0.8$ to get the mass of needed ore. At the end you obtain about $1.7$ tons of ore. Go!

To solve the second part, it is necessary to know the percentage of $\ce{CO}$ and $\ce{CO2}$ in the gas is a percentage in volume (= in amount of substances) or in mass.

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  1. What is the weight of the ore used per ton of pig iron?

The ore is $80\% \; \ce{Fe2O3}$, but the pig iron is $95\% \; \ce{Fe}$.

$$\pu{fraction}\ \ce{Fe}\ \pu{in} \; \ce{Fe2O3}= \dfrac{2\cdot56}{2\cdot56 + 3\cdot16}\times 100 = 0.70$$

$$\pu{tons ore} = \dfrac{95\%}{80\%\cdot0.70}= 1.7$$

  1. What is the volume of the blast furnace gas produced per ton of pig iron?

This solution to the second part is a bit more convoluted.

First we were given that "the ore is $80\% \; \ce{Fe2O3}$, $12\% \; \ce{SiO2}$, and [some fraction of] $\ce{Al2O3}$." For the components of the ore to add to $100\%$, there must be $8\% \ \ce{Al2O3}$.

Next the problem states "The blast furnace gas contains $28\% \ \ce{CO}$ and $12\% \ \ce{CO2}$." Here we must just assume that whatever the other 60% of the gas is, that it is inert. Also no temperature is given for the furnace so it is impossible to calculate a volume of gas, only a mass of $\ce{CO}$ can be calculated.

Now a twist here is that the carbon in the pig iron could have come from the $\ce{CO}$ or $\ce{CO2}$. However if the reaction

$$\ce{CO2 -> C + O2}\tag{2.1}$$

happened, then assuming complete combustion

$$\ce{2CO + O2 -> 2CO2}\tag{2.2}$$

Thus all of the carbon and all of the oxide removal is due to reactions with $\ce{CO}$.

$$\ce{Fe2O3 + 3CO <=> 2Fe + 3CO2 }\tag{2.3}$$ $$\ce{SiO2 + 2CO <=> Si + 2CO2}\tag{2.4}$$ $$\ce{Al2O3 + 3CO <=>2Al + 3CO2}\tag{2.5}$$ $$\ce{2CO <=> C + CO2}\tag{2.6}$$

The problem doesn't specify US tons or metric ton, so make it easy and assume metric ton. Now we need to figure out how much CO is needed, component by component.

\begin{array}{|c|c|c|c|}\hline Element & \%\ in \ pig \ iron & kg & kilo-moles & kilo-mole\ CO \\ \hline Fe &95\%& 950 & 950/56 = 17.0 & 3/2 \cdot 17.0 = 25.4\\ \hline Si &1.4\% &14 & 14/28 = 0.5 & 2\cdot 0.5 = 1.0 \\ \hline Al & - & & \\ \hline C & 3.6\% & 36 & 36/12 =3.0 &2\cdot 3 = 6\\ \hline \end{array}

Thus we need 32.4 kilo-moles of $\ce{CO}$ which is $28\cdot32.4 = \pu{907 kg}$, or 0.91 tons of $\ce{CO}$ per ton of pig iron (be they short tons, long tons, Imperial tons, or metric tons!)

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  • $\begingroup$ The volume of the gas has been asked. $\endgroup$
    – user586228
    Apr 18 '20 at 21:23
  • $\begingroup$ @user586228 - What is the temperature of the gas? As I noted since there is no temperature, it is impossible to calculate a volume. $\endgroup$
    – MaxW
    Apr 18 '20 at 21:25
  • $\begingroup$ Assume everything in Standard condition .Take 298K. $\endgroup$
    – user586228
    Apr 18 '20 at 21:26
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    $\begingroup$ 298K coming out of a blast furnace?!? // Anyway I've done the heavy lifting, I'll leave the volume calculation to you.Also if you want to do a volume calculation you'll have to assume some particular kind of ton. $\endgroup$
    – MaxW
    Apr 18 '20 at 21:28

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