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

$2 Na_3PO_4$ +$3 BaCl_2$ $\to$ 6NaCl + $B a_{3} (P O_{4})_{2}$

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Solution

The correct option is B
3

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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Similar questions

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 + $B a_{3} (P O_{4})_{2}$

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 + $B a_{3} (P O_{4})_{2}$

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{3 B a C l}_{2} + {2 N a}_{3} {P O}_{4} ⟶ {B a}_{3} {(P O}_{4})_{2} + 6 N a C l

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