Determine moles of limiting reagent In the reaction involving 0.1 g of calcium nitrate and 1.5 g of sodium phosphate, what is the number of moles of the limiting reagent?

Here is what I have so far:

Balanced equation:
$$\ce{3Ca(NO3)2 + 2Na3(PO4) -&gt; Ca3(PO4)2 + 6NaNO3}$$

$\ce{Ca(NO3)2}$ has MW = 164, mass = 0.1 g and determined moles using n=mass/MW = 0.000609
$\ce{Na3PO4}$ has MW = 164, mass = 1.5 g and determined moles using n=mass/MW = 0.0091496

I was able to identify limiting reagent by:
g $\ce{Ca(NO3)2}$ --> n $\ce{Ca(NO3)2}$ -->n $\ce{Na3(PO4)}$--> g $\ce{Na3(PO4)}$
    0.1 g --s1-> 0.000609 --s2--> 0.000406 -s3--> 0.06658 g

S1 = n = 0.1/164 = 0.000609
S2 = want/have x $\ce{Ca(NO3)2}$ = 2/3 x 0.000609 = 0.000406
S3 = m = nxMW = 0.000406 x 164 = 0.06658g

Therefore as all $\ce{Ca(NO3)2}$ did not use all the $\ce{Na3(PO4)2}$ that means the $\ce{Ca(NO3)2}$ is the limiting reagent.

Now to answer the question. What is the number of moles of the limiting reagent. I am not sure if it is 0.000609 OR 0.001828 (which is moles multiplied by the stoichiometric coefficient, which is 3).
It's just adding up to the existing answer since I don't have enough rep to comment. 0.000609 is the no. of moles of limiting reagent. You don't multiply it by it's stoichiometric co-efficient because it is the no. of moles you were given. Stoichiometric equations just signify the proportions in which compounds react.