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Levart [38]
3 years ago
10

How many grams of C are required to react with 97.1 g of Fe2O3?

Chemistry
1 answer:
sashaice [31]3 years ago
5 0

Answer:

21.9 g C

Explanation:

Fe₂O₃  +  3 C  ⇒  2 Fe  +  3 CO

This is your chemical equation.  

To find the grams of C required, first convert grams of Fe₂O₃ to moles using the molar mass (159.69 g/mol).  

(97.1 g)/(159.69 g/mol) = 0.608 mol Fe₂O₃

Then, use the mole ratio between Fe₂O₃ and C to convert moles of Fe₂O₃ to moles of C.  You can find the mole ratio by looking at the chemical equation.

(0.608 mol Fe₂O₃) × (3 mol Fe₂O₃/1 mol Fe₂O₃) = 1.824 mol C

You can then convert moles of C to grams using the molar mass (12.01 g/mol).

(1.824 mol) × (12.01 g/mol) = 21.9 g

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a) The conversion of 1-pentane to pentanoic acid using BH3, also known as hydroboration-oxidation, is a two-step reaction involving the reaction of 1-pentane with borane (BH3), followed by oxidation of the resulting 1-pentylborane with hydrogen peroxide or other oxidizing agents.

In the first step, 1-pentane reacts with borane (BH3) to form 1-pentylborane, through a process known as hydroboration. This reaction is catalyzed by a Lewis acid, such as aluminum chloride, and proceeds via a hydride transfer from the borane to the 1-pentane.

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The overall chemical reaction for the conversion of 1-pentane to pentanoic acid using borane (BH₃) and hydrogen peroxide (H₂O₂) is as follows:

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3. Protonation of the carbanion by water (or another proton source) to form pentanoic acid.

The overall reaction can be represented as follows:

1-Bromo butane + NaCN → Pentanoic Acid + NaBr

To know more about reagents, click below:

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