I don't think NiS and LiF will react. If we falsely assume that they will, this is the way to do the problem.
2LiF + NiS==> Li2S + NiF2
mols LiF = g/molar mass = 15.6/approx 33 = approx 0.48
mols NiS = approx 13/91 = approx 0.14
LiF will require 0.48/2 = 0.24 mols NiS. Do you have that much NiS? No you don't so NiS is the limiting reagent and LiF is the reagent in excess.
Not that my estimataes are just that. Even the atomic masses are estimates so you need to redo all of the calculations. Post your work if you get stuck.
15.6g of Lithium fluoride reacts with 13g of nickel (II) sulfide. What is the limiting reactant and the excessive reactant? Show work.
3 answers
Thank you! I need to know how many grams leftover of excess LiF there would be.
0.14 mols NiS will use up how much LiF. That will be
0.14 mols NiS will use twice that or approx 0.28 mols LiF. You had about 0.33 mols LiF initially so 0.33 - 0.28 = 0.05 approximately Lif mols in excess. Convert that to grams. g = mols x molar mass.\
Remember that this is a fantasy reaction. The reverse reaction will occur like this. Li2S + NiF2 ==> 2LiF + NiS but not the reaction you wrote initially. Also remember to redo all the calculations since I've estimated everything including atomic masses and molar masses.
0.14 mols NiS will use twice that or approx 0.28 mols LiF. You had about 0.33 mols LiF initially so 0.33 - 0.28 = 0.05 approximately Lif mols in excess. Convert that to grams. g = mols x molar mass.\
Remember that this is a fantasy reaction. The reverse reaction will occur like this. Li2S + NiF2 ==> 2LiF + NiS but not the reaction you wrote initially. Also remember to redo all the calculations since I've estimated everything including atomic masses and molar masses.
1 answer