70 tons of hematite (Fe₂O₃) is needed to produce 49 tons of iron (Fe).
We'll begin by calculating the tons of hematite (Fe₂O₃) that reacted and the tons of Fe produced from the balanced equation. This is illustrated below:
2Fe₂O₃ + 3C —> 4Fe + 3CO₂
Molar mass of Fe₂O₃ = (56×2) + (16×3) = 160 g/mol
Mass of Fe₂O₃ from the balanced equation = 2 × 160 = 320 g
Divide by 907185 to express in ton
320 / 907185 = 0.000353 ton
Molar mass of Fe = 56 g/mol
Mass of Fe from the balanced equation = 56 × 4 = 224 g
Divide by 907185 to express in ton
224 / 907185 = 0.000247 ton
From the balanced equation above,
0.000353 ton of Fe₂O₃ reacted to produce 0.000247 ton of Fe.
Finally, we shall determine the tons of hematite (Fe₂O₃) needed to produce 49 tons of Fe. This can be obtained as follow:
From the balanced equation above,
0.000353 ton of Fe₂O₃ reacted to produce 0.000247 ton of Fe.
Therefore,
X ton of Fe₂O₃ will react to produce 49 tons of Fe i.e
X ton of Fe₂O₃ = [tex]\frac{0.000353 * 49}{0.000247} \\\\[/tex]
X ton of Fe₂O₃ = 70 tons
Thus, 70 tons of hematite (Fe₂O₃) is needed to produce 49 tons of Fe.
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