homework help 1403

It’s a reaction.

##3 Hg(OH)_”2″ + 2 H_”3″PO_”4″ -> Hg_”3″(PO_”4″)_”2″ + 6 H_”2″O##

This is a neutralization reaction between a strong acid and a strong base.

First off, you missed a subscript of 3, in mercury (II) phosphate; should be ##Hg_”3″(PO_”4″)_”2″##

To balance it, you have to consider the subscripts and use coefficients effectively, so there is an equal amount of on either side of the equal sign.

##3 Hg(OH)_”2″ + 2 H_”3″PO_”4″ -> Hg_”3″(PO_”4″)_”2″ + 6 H_”2″O##

Hope this helps 🙂