PHS 100 - Self Assessment

Competency #7

Answer and Explanation

B₂O₃ (s) + HF (l) BF₃ (g) + H₂O (l)

To balance the equation, we must make sure there is the same number of each type of atom in the reactants as there is in the products. Rules for balancing the equation:

We are allowed to change the coefficients.
If no coefficient is shown, it has an implied value of ‘1'.
Only one coefficient per molecule/compound. The coefficient tells us how many of that molecule/compound are needed or are formed in the chemical reaction. For example ‘NH₃' means there is one NH₃ molecule (1 N atom and 3 H atoms), while ‘2 NH₃' means there are 2 NH₃ molecules (a total of 2 N atoms and 6 H atoms).
We are not allowed to change the subscripts while balancing the equation.

Make a table will help us to keep track of the number of each atom on each side:

B₂O₃ (s) + HF (1) BF₃ (g) + H₂O (1)

Reactants	Products	Reactants
2	B	1
3	O	1
1	H	2
1	F	3

We can see that the above chemical equation is not balanced.

First, we might try to change the coefficient in front of the BF₃ from ‘1’ to ‘2’ in an attempt to balance the boronn (B) atoms. Now, we update our table:

B₂O₃ (s) + HF (1) 2 BF₃ (g) + H₂O (1)

Reactants	Products	Reactants
2	B	2
3	O	1
1	H	2
1	F	6

We see that the equation is still not balanced.

Next, we try to get the flurine (F) atoms to balance by changing the coefficient in front of the HF from ‘1’ to ‘6’. Now, we update our table:

B₂O₃ (s) + 6 HF (1) 2 BF₃ (g) + H₂O (1)

Reactants	Products	Reactants
2	B	2
3	O	1
6	H	2
6	F	6

We see that the equation is still not balanced.

There are equal numbers of boron (B) and fluorine (F) on each side, but hydrogen (H) and oxygen (O) are not balanced. We can now change the coefficient for H₂O from ‘1’ to ‘3’. Again, we update our table:

B₂O₃ (s) + 6 HF (1) 2 BF₃ (g) + 3 H₂O (1)

Reactants	Products	Reactants
2	B	2
3	O	3
6	H	6
6	F	6

Now, the equation is balanced! Often, it will take some trial and error to get a balanced chemical equation.

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