# Question #6837f

Mar 29, 2016

The carbonate is provided by the $N {a}_{2} C {O}_{3}$ and the magnesium from the $M g {\left(N {O}_{3}\right)}_{2} . 6 {H}_{2} O$. ONLY the Mg and $C {O}_{3}$ in the final product matter. First, a balanced equation is always required.
$M g {\left(N {O}_{3}\right)}_{2} . 6 {H}_{2} O$ + $N {a}_{2} C {O}_{3}$$M g C {O}_{3} \left(s\right) + 2 N {O}_{3}^{1 -} + 2 N {a}^{1 +}$ (water is in solution)
Then calculate the number of moles in 2.00g $M g C {O}_{3}$. 2.00/84.3 = 0.0237. This is then also the total number moles of Mg and $C {O}_{3}$ used. Whichever reactant was present in larger quantity (no quantities were specified in the question) is the excess reagent. The other one is the limiting reagent.