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.
Show 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.
Step 1: Understanding the Concept:
We need the mass of ore that, after losing some iron to slag and being reduced from $Fe_2O_3$ to metallic Fe, ends up giving the iron content of $1000$ kg of hot metal.
Step 2: Key Formula or Approach:
Build one chain: ore mass to $Fe_2O_3$ mass to iron charged to iron surviving to hot metal, matching the Fe needed in $1000$ kg hot metal. Write it as a single equation and solve for ore mass $W$.
\[ W \times 0.85 \times \frac{112}{160} \times 0.98 = Fe_{needed} \]
Step 3: Detailed Explanation:
Since hot metal is $4\%$ C and $1.5\%$ Si, the iron share is
\[ Fe_{needed} = 1000\times(1-0.04-0.015) = 945 \text{ kg} \]
The combined yield factor from ore to hot metal iron is
\[ 0.85 \times \frac{112}{160} \times 0.98 = 0.85\times0.70\times0.98 = 0.5831 \]
This single number tells us that out of every kg of ore, only $0.5831$ kg ends up as iron in the hot metal.
Step 4: Final Answer:
Solving for $W$,
\[ W = \frac{945}{0.5831} = 1620.6 \text{ kg} \]
So about $1620.6$ kg of ore is needed per $1000$ kg of hot metal.
\[ \boxed{W \approx 1620.6 \text{ kg}} \]
