Limiting reactant problem.
There are several ways to do limiting reactant problems, but the easiest way to identify the limiting reactant, in my opinion, is to find the moles of each reactant, divide each by the respective coefficient in the balanced equation, and whichever is less is the limiting reactant. For this problem, we proceed as follows:
Write a correctly balanced equation:
Ca(NO3)2(aq) + Na2CO3(aq) ==> CaCO3(s) + 2NaNO3(aq)
Find moles of each and divide by coefficient:
moles Ca(NO3)2 = 20.00 ml x 1 L/1000 ml x 0.190 mol/L = 0.0038 mol (÷ 1 = 0.0038)
moles Na2CO3 = 12.00 ml x 1 L/1000 ml x 0.150 mol/L = 0.0018 mol (÷ 1 = 0.0018)
Since 0.0018 is less than 0.0038 Na2CO3 is limiting.
(NOTE: in this case it was easy to see what is limiting because mole ratio is 1:1)
The longer way to do the same thing is to determine the moles of product (any product) formed from each reactant, and see which one produces the smaller amount. This would be as follows:
moles CaCO3 produced = 20.00 ml Ca(NO3)2 x 1L/1000 ml x 0.190 mol/L x 1 mol CaCO3/mol Ca(NO3)2 = 0.0038 moles CaCO3 produced
moles CaCO3 produced = 12.00 ml x 1L/1000 ml x 0.15 mol/L x 1 mol CaCO3/mol Na2CO3 = 0.0018 moles CaCO3 produced. Thus, Na2CO3 is limiting.