# Question #53720

##### 1 Answer
Oct 3, 2015

$3 M g {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + 2 N {a}_{3} P {O}_{4 \left(a q\right)} \to M {g}_{3} {\left(P {O}_{4}\right)}_{2 \left(s\right)} + 6 N a N {O}_{3 \left(a q\right)}$

#### Explanation:

This is a double displacement reaction, which means that the cation of the first compound will couple with the anion of the second and the cation of the second compound will couple with the anion of the first.

The reaction would be:

$3 M g {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + 2 N {a}_{3} P {O}_{4 \left(a q\right)} \to M {g}_{3} {\left(P {O}_{4}\right)}_{2 \left(s\right)} + 6 N a N {O}_{3 \left(a q\right)}$