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sergeinik [125]
3 years ago
12

Methyl bromide (CH3Br) is produced naturally by fungi. Methyl bromide has also been used in agriculture as a fumigant, but this

use is being phased out because the compound has been linked to ozone depletion in the atmosphere. Draw the Lewis structure of CH3Br with the formal charges minimized. Include lone pairs of electrons.

Chemistry
1 answer:
Troyanec [42]3 years ago
8 0

Answer:

See image attached for structure of CH3Br

Explanation:

A lewis structure structure is a representation of a covalent compound in which dots are used to show valence electrons, lone pairs and bonding electrons. The system was introduced by sir G.N Lewis in 1916 in his article titled 'Atom and Molecule'. They are also called dot electron diagrams of molecules. CH3Br molecule contains a total of fourteen electrons. Valence electrons are shown by dots around the atom of each element as clearly seen in the image attached.

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1.86 g H2 is allowed to react with 9.75 g N2 , producing 2.87g NH3.
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                     (a)  Theoretical Yield  =  10.50 g

                      (b)   %age yield  = 27.33 %

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Answer-Part-(a)

                 The balance chemical equation for the synthesis of Ammonia is as follow;

                                          N₂ + 3 H₂ → 2 NH₃

Step 1: Calculating moles of N₂ as;

                   Moles = Mass / M/Mass

                   Moles = 9.75 g / 28.01 g/mol

                   Moles = 0.348 moles of N₂

Step 2: Calculating moles of H₂ as;

                   Moles = Mass / M/Mass

                   Moles = 1.86 g / 2.01 g/mol

                   Moles = 0.925 moles

Step 3: Finding Limiting reagent as;

According to equation,

                1 mole of N₂ reacts with  =  3 moles of H₂

So,

             0.348 moles of N₂ will react with  =  X moles of H₂

Solving for X,

                     X = 3 mol × 0.348 mol / 1 mol

                     X = 1.044 mol of H₂

It shows that to consume 0.348 moles of N₂ completely we require 1.044 mol of Hydrogen while, as given in statement we are only provided with 0.925 moles of H₂ hence, hydrogen  is limiting reagent. Therefore, H₂ will control the final yield.

Step 4: Calculating moles of Ammonia as,

According to equation,

                3 mole of H₂ produces  =  2 moles of NH₃

So,

             0.925 moles of H₂ will produce  =  X moles of NH₃

Solving for X,

                     X = 2 mol × 0.925 mol / 3 mol

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Step 5: Calculating theoretical yield of Ammonia as,

                     Theoretical Yield  =  Moles × M.Mass

                     Theoretical Yield  =  0.616 mol  × 17.03 g/mol

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                    %age yield  = 2.87 g / 10.50 g × 100

                    %age yield  = 27.33 %

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