Q6). Balance the following chemical equations

$$ (a) \;\;\; HNO_{3} + Ca(OH)_{2} $$

$$ \downarrow$$

$$ \;\;\; Ca(NO_{3})_{2} + H_{2}O $$

$$ (b) \;\;\; NaOH + H_{2}SO_{4} $$

$$ \downarrow$$

$$ \;\;\; Na_{2}SO_{4} + H_{2}O $$

$$ (c) \;\;\; NaCl + AgNO_{3} $$

$$ \downarrow$$

$$ \;\;\; AgCl + NaNO_{3} $$

$$ (d) \;\;\; BaCl_{2} + H_{2}SO_{4} $$

$$ \downarrow$$

$$ \;\;\; BaSO_{4} + HCl $$

a) Let's count the number of atoms of each element in reactants and products side.
On reactants side:
H:3,N:1,O:5,Ca:1
On products side:
H:2,N:2,O:7,Ca:1
By Hit and trial method
Let's try to balance N by multiplying 2 to $$ HNO_{3} $$ on reactants side.
Now we have:
On reactants side: H:4,N:2,O:8,Ca:1
On products side: H:2,N:2,O:7,Ca:1
Now if we multiply 2 to $$H_{2}O$$ on products side
Equation so far :

$$ \;\;\; 2HNO_{3} + Ca(OH)_{2} $$

$$ \downarrow$$

$$ \;\;\; Ca(NO_{3})_{2} + 2H_{2}O $$

Since the numbers on the left equal the right, you now have a balanced equation, so this must be the answer:

SIMILARLY.

$$ (b) \;\;\; 2NaOH + H_{2}SO_{4} $$

$$ \downarrow$$

$$ \;\;\; Na_{2}SO_{4} + H_{2}O $$

$$ (c) \;\;\; NaCl + AgNO_{3} $$

$$ \downarrow$$

$$ \;\;\; AgCl + NaNO_{3} $$

$$ (d) \;\;\; BaCl_{2} + H_{2}SO_{4} $$

$$ \downarrow$$

$$ \;\;\; BaSo_{4} + 2HCl $$