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: $$\ce{3Ca(NO3)2 + 2Na3(PO4) -> Ca3(PO4)2 + 6NaNO3}$$

$\ce{Ca(NO3)2}$ 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).

  • $\begingroup$ Welcome to Chemistry! This seems like a homework question. We ‎have a policy which states that you should show your thoughts and/or efforts into solving the ‎problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this ‎question may get closed. $\endgroup$ – bon Mar 17 '16 at 9:19
  • $\begingroup$ Sorry this would be my first post on a forum. I have edited what i have so far under the question. Hope you can clarify my query! $\endgroup$ – user27980 Mar 17 '16 at 12:25
  • $\begingroup$ Ok, please excuse my ignorance as mentioned above this is my first thread on a forum. I believe i have fulfilled the requirements of the question type to allow it to be on-topic. Can anyone assist me in my confusion? $\endgroup$ – user27980 Mar 17 '16 at 14:07
  • $\begingroup$ You can already see from your first calculation that Ca(No3)2 is the limiting reagent, because you had more moles of Na3PO4 than Ca3(No3)2 and the equation tells us that 3 moles of Ca3(No3)2 react with 2 moles of Na3PO4. How much did exactly react? 0.000609 moles Ca3(NO3)2, 2/3 x 0.000609 moles Na3PO4 and the amount of moles NaPO4 left after the reaction: 0.0091496 - (2/3 x 0.000609). $\endgroup$ – user21398 Mar 17 '16 at 17:14
  • $\begingroup$ user21398 I understand exactly what you have said BUT I am still confused! Would the solution to this problem of "What is the number of moles of the limiting reagent?" be 0.000609 OR 0.001828 $\endgroup$ – user27980 Mar 17 '16 at 20:27

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.


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