Calculate n-factor of N a 3 P O 4 in the following reaction. N a 3 P O 4 + 3 B a C l 2 → 6NaCl + B a 3 ( P O 4 ) 2 Options: A . &nbsp 2 B . &nbsp 3 C . &nbsp 4 D . &nbsp 6

Calculate n-factor of Na3PO4 in the following reaction. Na3PO4 +3B

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Question -

Calculate n-factor of $N a_{3} P O_{4}$ in the following reaction.

$N a_{3} P O_{4}$ + $3 B a C l_{2}$ $\to$ 6NaCl + $B a_{3} (P O_{4})_{2}$

Options:

A . 2

B . 3

C . 4

D . 6

Answer: Option B
:
B

This is the type of reaction where none of the atoms involved undergo a change in their oxidation states.
A simple way to find the n-factor of $N a_{3} P O_{4}$ in such reactions would be to find the amount of cationic charge replaced per molecule.
In $2 N a_{3} P O_{4}$ , we have ${6 N a}^{+}$
So, $N a_{3} P O_{4}$ has ${3 N a}^{+}$

${N a}^{+}$ can be cosider to be similar to $H^{+}$ in $H_{3} P O_{4}$ (just an analogy)

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