Question

Calculate n-factor of $N{a}_{3}P{O}_4$ in the following reaction.

$N{a}_{3}P{O}_4$ +$3BaC{l}_2$ $\to$ 6NaCl + $B{a}_3(P{O}_4{)}_2$

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

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 $2N{a}_{3}P{O}_4$, we have${6Na}^{+}$
So, $N{a}_{3}P{O}_4$ has ${3Na}^{+}$

${Na}^{+}$ can be cosider to be similar to $H^{+}$in $H_{3}P{O}_4$ (just an analogy)