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balandron [24]
3 years ago
6

How many monochlorination products can be formed (constitutional isomers only) from the reaction of (CH3)2CHCH2CH3 with Cl2 and

hν?
Chemistry
1 answer:
faltersainse [42]3 years ago
7 0

Answer:

4 monochlorination products can be formed.

Explanation:

Constitutional isomers : These are those compounds with same molecular formula but different atomic arrangement.For example: butane and 2-methly-propane.

On monochlorination of 2 methyl-butane  we will have four possibilities of product which will be constitutional isomers of each other:

  • 1-chloro-3-methyl butane : (CH_3)_2CH-CH_2-CH_2-Cl
  • 2-chloro-3-methyl butane : (CH_3)_2CH-CHCl-CH_3
  • 2-chloro-2-methyl butane (CH_3)_2CCl-CH_2-CH_3
  • 1-chloro-2-methyl butane (CH_2Cl)_2CH-CH_2-CH_3
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Answer:

After complete reaction, 0.280 moles of ammonia are produced

Explanation:

Step 1: Data given

Number of moles N2 = 0.140 moles

Number of moles H2 = 0.434 moles

Step 2: The balanced equation

N2(g) + 3H2 (g) ⟶ 2NH3 (g)

Step 3: Calculate the limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. It will completely be consumed (0.140 moles).

H2 is in excess. There will react 3*0.140 = 0.420 moles

There will remain 0.434 - 0.420 = 0.014 moles

Step 4: Calculate moles NH3

For 0.140 moles N2 we'll have 2*0.140 = 0.280 moles NH3

After complete reaction, 0.280 moles of ammonia are produced

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tresset_1 [31]

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