Question #6837f

1 Answer
SCooke
Mar 29, 2016

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.

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