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Pavel [41]
3 years ago
9

Balance 6Sb +6Cl2– 6SbCl3

Chemistry
1 answer:
labwork [276]3 years ago
4 0

<u>Answer:</u>

<u>Balancin</u>g is making the number of atoms of each element same on both the sides  (reactant and product side).

To find the number of atoms of each element we multiply coefficient and the subscript  

For example 5 Ca_1 Cl_2 contains

5 × 1 = 5,Ca atoms and

5 × 2 = 10, Cl atoms

If there is a bracket in the chemical formula

For example 3Ca_3 (P_1 O_4 )_2 we multiply coefficient \times subscript \times number outside the bracket.......... to find the number of atoms  

(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )

So

3 × 3 = 9, Ca atoms  

3 × 1 × 2 = 6, P atoms  

3 × 4 × 2 = 24, O atoms are present.

So

Let us balance the equation given

1Sb +1Cl_2 > 1SbCl_3

(Unbalanced)

Reactant side - Number of atoms of each element - Product side

1  - Sb - 1

2 - Cl - 3  

So we have on the product side odd number of Cl to convert to even number, multiply SbCl_3 by 2

The equation changes to

1Sb +1Cl_2> 2SbCl_3

Reactant side - Number of atoms of each element - Product side

1  - Sb - 2

2 - Cl - 6

Multiplying Sb by 2 and Cl_2

2Sb +3Cl_2> 2SbCl_3

The equation is balanced now!!!!  

Reactant side - Number of atoms of each element - Product side

2  - Sb - 2

6 - Cl - 6

Balanced.

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Answer:

Synthesis

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An intermediate in this reaction is seleninyl fluoride (SeOF2).

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Structure and bonding

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