Question

Given the equation: __ca 2+ (aq) + __po43− (aq) → __ca3(po4)2(s) when the equation is correctly balanced, the sum of the total charge of the reactants is

Answer 1

`3Ca^(2+)`(aq) +
`2P O_(4) ^(3-)` (aq) ⇒
`Ca_(3) (P O_(4) )_(2) (s)`


The sum of total charges of the reactant = 3×(2+) + 2×(3-)
= 6 + (-6)
=0