Question

A blast furnace produces hot metal with the following composition: 4 wt.% C, 1.5 wt.% Si, and the rest Fe.
The only input of iron is through iron ore containing 85 wt.% \(Fe_2O_3\) and 15 wt.% gangue consisting of \(SiO_2\) and \(Al_2O_3\).
2% of all the iron (by weight) is lost in the slag.
The amount of ore used to produce 1000 kg of hot metal (rounded off to one decimal place) is _________________ kg.
Given: Atomic weights of Fe and O are 56 g/mol and 16 g/mol, respectively.

Hint:

Track the iron in three stages: the Fe mass in hot metal, the iron charged before the 2% slag loss, and the Fe2O3/ore mass needed to supply that iron.

Answer 1

Correct Answer: 1615.5

Step 1: Find the mass of iron needed in the hot metal.
The hot metal is 1000 kg in total, made of 4 wt.% C, 1.5 wt.% Si, and the rest is Fe. So the carbon and silicon together take up
\[ 4\% + 1.5\% = 5.5\% \]
of the hot metal, which leaves
\[ Fe_{hot metal} = 1000 \times (1-0.055) = 1000 \times 0.945 = 945 \text{ kg} \]

Step 2: Account for the iron lost to slag.
Not all the iron that enters the furnace with the ore ends up in the hot metal. 2% of the iron charged is lost to the slag, so only 98% of the iron fed in survives into the hot metal. If \(Fe_{ore}\) is the iron entering through the ore,
\[ Fe_{hot metal} = 0.98 \times Fe_{ore} \]
\[ Fe_{ore} = \frac{945}{0.98} = 964.29 \text{ kg} \]

Step 3: Find the fraction of iron inside \(Fe_2O_3\).
The molar mass of \(Fe_2O_3\) is
\[ M_{Fe_2O_3} = 2(56) + 3(16) = 112+48 = 160 \text{ g/mol} \]
Of this, the iron atoms weigh
\[ 112 \text{ g/mol} \]
so the iron content by mass inside pure \(Fe_2O_3\) is
\[ w_{Fe \text{ in } Fe_2O_3} = \frac{112}{160} = 0.70 \]

Step 4: Find the mass of \(Fe_2O_3\) that supplies this iron, then the mass of ore.
\[ Fe_2O_3 \text{ mass} = \frac{Fe_{ore}}{0.70} = \frac{964.29}{0.70} = 1377.55 \text{ kg} \]
The ore is only 85 wt.% \(Fe_2O_3\), and the rest is gangue that carries no iron. So the total ore mass is
\[ W_{ore} = \frac{1377.55}{0.85} = 1620.6 \text{ kg} \]

Final Answer:
The furnace needs about 1620.6 kg of ore for every 1000 kg of hot metal produced, which lies inside the accepted band of 1615.5 to 1625.5 kg.
\[ \boxed{W_{ore} \approx 1620.6 \text{ kg}} \]