Answer:
a) ![HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O](https://tex.z-dn.net/?f=HClO_4%20%28aq%29%20%2B%20LiOH%20%5Crightarrow%20LiClO_4%20%2B%20H_2O)
b) ![H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)](https://tex.z-dn.net/?f=H_2SO_4%28aq%29%20%2B%202NaOH%28aq%29%20%5Crightarrow%20Na_2SO_4%20%28aq%29%20%2B%202H_2O%20%28l%29)
c) ![Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O](https://tex.z-dn.net/?f=Ba%28OH%292%20%2B%202HF%5Crightarrow%20BaF_2%20%2B%202H_2O)
Explanation:
a)
when
is added to LiOH, lithium chlorate and water is formed.
![HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O](https://tex.z-dn.net/?f=HClO_4%20%28aq%29%20%2B%20LiOH%20%5Crightarrow%20LiClO_4%20%2B%20H_2O)
Balancing of above reaction,
It can be seen that all the atoms in both the sides are balanced. so it is a balanced reaction.
(b) When aqueous
is added to NaOH,
and
is formed. It is a neutrilization reaction.
![H_2SO_4(aq) + NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)](https://tex.z-dn.net/?f=H_2SO_4%28aq%29%20%2B%20NaOH%28aq%29%20%5Crightarrow%20Na_2SO_4%20%28aq%29%20%2B%20H_2O%20%28l%29)
Balancing of above reaction,
First balance all the atoms except O and H
S atom is already balanced in either side
No. of Na atom in left hand side = 1
No. of Na atom in right hand side = 2
So multiply NaOH by 2, now the reaction becomes:
![H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)](https://tex.z-dn.net/?f=H_2SO_4%28aq%29%20%2B%202NaOH%28aq%29%20%5Crightarrow%20Na_2SO_4%20%28aq%29%20%2B%20H_2O%20%28l%29)
No. of O atoms in left hand side = 6
No. of O atoms in right hand side = 5
So multiply H2O by 2, now the reaction becomes:
![H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)](https://tex.z-dn.net/?f=H_2SO_4%28aq%29%20%2B%202NaOH%28aq%29%20%5Crightarrow%20Na_2SO_4%20%28aq%29%20%2B%202H_2O%20%28l%29)
Now, it can been seen that all the atoms are balanced.
c) When
reacts with HF, baroum fluoride and water is formed.
![Ba(OH)2 + HF\rightarrow BaF_2 + H_2O](https://tex.z-dn.net/?f=Ba%28OH%292%20%2B%20HF%5Crightarrow%20BaF_2%20%2B%20H_2O)
Balancing of above reaction,
Barium atoms are already balanced on either side.
No. of F atoms on left hand side = 1
No. of F atoms on right hand side = 2
So, multiply HF by 2, now reaction becomes
![Ba(OH)2 + 2HF\rightarrow BaF_2 + H_2O](https://tex.z-dn.net/?f=Ba%28OH%292%20%2B%202HF%5Crightarrow%20BaF_2%20%2B%20H_2O)
In order to balance H and O on either side, multiply H2O by 2.
![Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O](https://tex.z-dn.net/?f=Ba%28OH%292%20%2B%202HF%5Crightarrow%20BaF_2%20%2B%202H_2O)