Question #6837f

1 Answer
Mar 29, 2016

Answer:

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.