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Rudik [331]
3 years ago
9

This question has multiple parts. work all the parts to get the most points. one method for preparing a nitrile is the dehydrati

on of a primary amide. provide the mechanism for the reaction below: reaction a use curved arrows to show the mechanism of the step below. make the ends of your arrows specify the origin and destination of reorganizing electrons. arrow-pushing instructions b use curved arrows to show the mechanism of the step below. make the ends of your arrows specify the origin and destination of reorganizing electrons. arrow-pushing instructions c use curved arrows to show the mechanism of the step below. make the ends of your arrows specify the origin and destination of reorganizing electrons. arrow-pushing instructions d use curved arrows to show the mechanism of the step below. make the ends of your arrows specify the origin and destination of reorganizing electrons. arrow-pushing instructions

Chemistry
1 answer:
nikdorinn [45]3 years ago
4 0
Following reaction shows the preparation of Nitrile from Amide. In this reaction Thionyl Chloride (SOCl₂)is reacted with Amide, which produced nitrile along with SO₂ and HCl gas. The importance of this reaction is the removal of byproducts (i.e SO₂ and HCl). As both SO₂ and HCl are gases so no extra workup is required to remove them. The mechanism of reaction is as follow,

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Answer:

Explanation:

  • For the balanced reaction:

<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)​.</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

  • Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

  • Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

  • Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

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The rocket is now too heavy to reach its destination.

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