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# Balance the following chemical equations $$(a) \;\;\; HNO_{3} + Ca(OH)_{2}$$ $$\downarrow$$$$\;\;\; Ca(NO_{3})_{2} + H_{2}O$$ $$(b) \;\;\; 2NaOH + H_{2}SO_{4}$$ $$\downarrow$$$$\;\;\; Na_{2}SO_{4} + 2H_{2}O$$ $$(c) \;\;\; NaCl + AgNO_{3}$$ $$\downarrow$$$$\;\;\; AgCl + NaNO_{3}$$ $$(d) \;\;\; BaCl_{2} + H_{2}SO_{4}$$ $$\downarrow$$$$\;\;\; BaSO_{4} + 2HCl$$

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.

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

$$\downarrow$$

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

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

$$\downarrow$$

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

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

$$\downarrow$$

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