this is all my work so far on the limiting reagent problem I asked about yesterday about Calcium nitrate and ammonium fluoride....

28.42g/164.1= .17318 mol Ca(NO3)2

29.6g/37.042=.79909 mol (NH4)F

(1mol Ca(NO3)2 /.17318) x (2mol (NH4)F/x)=.34636

(2 mol (NH4)F/.79909)x(1 mol Ca(NO3)2/x)=.399545

so (NH4)F (.34636) is limiting reagent.

now I'm stuck. Not sure what you mean about subtracting moles from initial moles etc.....thanks!

this is all my work so far on the limiting reagent problem I asked about yesterday about Calcium nitrate and ammonium fluoride....

28.42g/164.1= .17318 mol Ca(NO3)2
OK

29.6g/37.042=.79909 mol (NH4)F
OK

(1mol Ca(NO3)2 /.17318) x (2mol (NH4)F/x)=.34636 moles NH4F needed AND YOU HAVE THAT MUCH so NH4F is not the limiting reagent.

(2 mol (NH4)F/.79909)x(1 mol Ca(NO3)2/x)=.399545 moles Ca(NO3)2 needed and you DON'T have that much; therefore, Ca(NO3)2 is the limiting reagent.

so (NH4)F (.34636) is limiting reagent.

now I'm stuck. Not sure what you mean about subtracting moles from initial moles etc.....thanks!
Since Ca(NO3)2 is the limiting reagent, all of it will be used and there will be none remaining after the reaction is complete.
How much NH4F remains? From your calculation above, all of the Ca(NO3)2 (0.1732 moles) requires 0.3464 moles NH4F so 0.7991 mole NH4F initially - 0.3464 moles reacted = 0.4527 moles NH4F that remains unreacted. Multiply by molar mass NH4F to obtain grams NH4F unreacted. I don't remember if the original problem asked for the amount of products or not. If not then this is the end of the problem. If so, you can post again if you have trouble getting those numbers.

I re-read the original problem. The way I read it the problem asks for the products as well so use the limiting reagent amount of Ca(NO3)2 for your calculations of the moles CaF2, N2O, and H2O produced, then change from moles to grams. Post again if you get stuck.

Thank you so much DrBobb for all of your help!!

Great job on finding the moles of calcium nitrate and ammonium fluoride! Now, let's move on to determining the limiting reagent and calculating the excess reagent.

To determine the limiting reagent, we compare the mole ratio of the reactants to the balanced equation. In this case, we have:

2 Ca(NO3)2 + 4 (NH4)F -> 2 CaF2 + 4 NH4NO3

From your calculations, you found that there are 0.34636 moles of (NH4)F and 0.17318 moles of Ca(NO3)2. The mole ratio of (NH4)F to Ca(NO3)2 is 4:2 or simply 2:1, as per the balanced equation.

So, for every 2 moles of (NH4)F, we need 1 mole of Ca(NO3)2. Since we have less Ca(NO3)2 (0.17318 moles) than the ratio requires, it is the limiting reagent. This means that ammonium fluoride is in excess.

To find the excess reagent, we need to determine how many moles of the excess reagent is left after the reaction with the limiting reagent has occurred. In this case, the limiting reagent is Ca(NO3)2.

To find the amount of excess (NH4)F, we subtract the moles of (NH4)F used in the reaction from the initial moles. Since we have 0.34636 moles of (NH4)F and it is in excess, there is no (NH4)F left. Therefore, the moles of excess (NH4)F is 0.

Please note that this is a simplified explanation of the limiting reagent problem. In actual scenarios, you should also consider the stoichiometry and actual masses of the reactants.

Keep up the great work! Let me know if you have any further questions.