Question

Question #6837f

Answer 1

The limiting reagent is the one that was only 0.237 moles. The other one is the excess reagent.

Explanation:

The carbonate is provided by the `Na_2CO_3` and the magnesium from the `Mg(NO_3)_2 . 6H_2O`. ONLY the Mg and `CO_3` in the final product matter. First, a balanced equation is always required.

`Mg(NO_3)_2 . 6H_2O` + `Na_2CO_3` → `MgCO_3 (s) + 2NO_3^(1-) + 2Na^(1+)` (water is in solution)

Then calculate the number of moles in 2.00g `MgCO_3`. 2.00/84.3 = 0.0237. This is then also the total number moles of Mg and `CO_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.