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