2Na3PO4 + 3CaCl2 ----> Ca3 (PO4)2 + 6NaCl

how many moles of CaCl2 remain if .10 mol Na3PO4 and .40 mol CaCl2 are used?

When amounts of both materials are listed one must worry about which is the limiting reagentalthough the problem seems to suggest that Na3PO4 is the limiting reagent. We can take them in order. I don't have great eye sight but I think your numbers are 0.10 mol Na3PO4 and 0.40 mol CaCl2.

0.10 mol Na3PO4 x [3 mol CaCl2/2 mol Na3PO4] =0.15 mol CaCl2 and that subtracted from 0.40 = 0.25 mol CaCl2 remaining. You can check this out; all of the Na3PO4 should have been used and it is the limiting reagent.

0.40 mol CaCl2 x [2 mol Na3PO4/3 mol CaCl2 = 0.267 mol Na3PO4. We didn't have that much initially; therefore, all of it must have been consumed. I hope this helps.

0.25

Well, it seems like you already got the answer! According to your calculations, there will be 0.25 moles of CaCl2 remaining. So, to answer your question directly, 0.25 moles of CaCl2 will remain if 0.10 moles of Na3PO4 and 0.40 moles of CaCl2 are used. Just remember, though, those are actual numbers. If you were looking for a humorous answer, here it goes: "Well, it looks like the CaCl2 is a bit clingy and doesn't want to let go. It's holding on to 0.25 moles for dear life!" Keep in mind, though, that the actual explanation you provided is pretty spot-on!

To find out how many moles of CaCl2 remain, we need to determine the limiting reagent first.

From the balanced equation:
2Na3PO4 + 3CaCl2 -> Ca3(PO4)2 + 6NaCl

We can see that the stoichiometric ratio between Na3PO4 and CaCl2 is 2:3. This means that for every 2 moles of Na3PO4, we need 3 moles of CaCl2.

Given that there are 0.10 mol Na3PO4 and 0.40 mol CaCl2, we can calculate how many moles of CaCl2 are required based on the amount of Na3PO4 used.

0.10 mol Na3PO4 x (3 mol CaCl2 / 2 mol Na3PO4) = 0.15 mol CaCl2

So, 0.15 mol CaCl2 would be required to react completely with 0.10 mol of Na3PO4.

Since we initially have 0.40 mol of CaCl2, we can find the number of moles remaining by subtracting the amount used from the initial amount:

0.40 mol CaCl2 - 0.15 mol CaCl2 = 0.25 mol CaCl2

Therefore, 0.25 mol of CaCl2 would remain after the reaction.

To determine the amount of CaCl2 remaining, you need to compare the amount of CaCl2 used to the amount of Na3PO4 used.

First, convert the given moles of Na3PO4 to moles of CaCl2 using the stoichiometry of the balanced equation:

0.10 mol Na3PO4 x (3 mol CaCl2 / 2 mol Na3PO4) = 0.15 mol CaCl2

This means that 0.15 mol of CaCl2 was used.

To find the amount of CaCl2 remaining, subtract the amount used from the initial amount:

0.40 mol CaCl2 - 0.15 mol CaCl2 = 0.25 mol CaCl2

Therefore, 0.25 mol of CaCl2 remains after the reaction.

Idiot above me is wrong