# Dr.Bob222

posted by on .

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!

• Dr.Bob222 - ,

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.

• Dr.Bob222 - ,

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.

• Dr.Bob222 - ,

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