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Answer :
Final answer:
The amount of charcoal (carbon) required to reduce 40kg of iron(III)oxide (Fe2O3) is approximately 9.0 kg, which does not match any of the options given (37.6 kg, 40.0 kg, 53.3 kg, 80.0 kg). There could be a mistake in the options or in the original problem.
Explanation:
To calculate the amount of charcoal (carbon) needed to reduce 40 kg of iron(III)oxide (Fe2O3), you need to know the balanced chemical equation for this reduction process:
Fe2O3 + 3C → 2Fe + 3CO2.
From this balanced equation, we see that we need 3 moles of carbon (C) to fully reduce 1 mole of Fe2O3 to iron (Fe). The molar mass of Fe2O3 is about 159.69 g/mol and of C is about 12.01 g/mol. Given this, if we have 40,000 g (which is 40 kg) of Fe2O3, that's approximately 250.4 moles (40,000/159.69).
So, we'll require 3 times this amount in moles of Carbon, which is about 751.2 moles (3 * 250.4). Now converting this moles of Carbon to mass using its molar mass, it gives us 9,015.6 g or roughly 9.0 kg (751.2 * 12.01). This is not matching any of the options provided which are a) 37.6 kg b) 40.0 kg c) 53.3 kg d) 80.0 kg.
There seems to be some mistake in the options provided, or possibly in the original problem itself. If you are certain about the data provided in the problem, you may want to double check with your teacher or textbook.
Learn more about Stoichiometry here:
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