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

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

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

2

N a_{3} P O_{4}

B a C l_{2}

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

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

B a C l_{2}

N a_{3} P O_{4},

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

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

0.3

B a C l_{2}

0.2

N a_{3} P O_{4}

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

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_{2}

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