In the following reaction, the values of a, b and c, respectively are
a F 2 (g) + b OH – (aq) ⎯→ c F – (aq) + d OF 2 (g) + e H 2 O(l)
Text Solution
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1. Identify the elements involved and their initial counts:
Fluorine ( 
)
Oxygen (O)
Hydrogen 
2. Write the unbalanced equation:
3. Balance the fluorine atoms:
There are 2 fluorine atoms on the reactant side in 
.
There are fluorine atoms in both 
and 
on the product side.
Let's balance the fluorine atoms by assuming:
, so we have 
, giving 4 fluorine atoms on the reactant side.
Assume 
(4 fluoride lons), leaving 2 fluorine atoms for 
(1 molecule of 
).
So, we get:
4. Balance the oxygen and hydrogen atoms:
On the reactant side, there is 1 oxygen atom in 
.
On the product side, there is 1 oxygen atom in 
and 1 oxygen atom in 
.
We balance the oxygen atoms by assuming 
( 2 hydroxide ions):
5. Verify the hydrogen balance:
On the reactant side, we have 2 hydrogen atoms in 
.
On the product side, we need 2 hydrogen atoms, which are provided by 1 molecule of 
The balanced equation is:
Therefore, the values of 
, and 
are 2,2, and 4, respectively.
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