15.6g of Lithium fluoride reacts with 13g of nickel (II) sulfide. What is the limiting reactant and the excessive reactant? Show work.
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.
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.
To determine the limiting reactant and excessive reactant, we need to compare the number of moles for each reactant and see which one is limiting the reaction.
First, we need to find the molar mass of each compound:
- Lithium fluoride (LiF) has a molar mass of 25.94 g/mol (lithium) + 18.99 g/mol (fluorine) = 44.93 g/mol
- Nickel (II) sulfide (NiS) has a molar mass of 58.69 g/mol (nickel) + 32.07 g/mol (sulfur) = 90.76 g/mol
Next, we can calculate the number of moles for each compound using the given mass and molar mass:
- Moles of LiF = mass of LiF / molar mass of LiF = 15.6 g / 44.93 g/mol = 0.347 mol
- Moles of NiS = mass of NiS / molar mass of NiS = 13 g / 90.76 g/mol = 0.143 mol
Now, we compare the ratio of moles of each compound to their stoichiometric coefficients in the balanced chemical equation. The balanced chemical equation is necessary for this step, so it would be helpful to have it.
Let's assume the balanced chemical equation for the reaction is:
2 LiF + NiS -> Li2S + NiF2
From the balanced equation, we can see that the ratio between LiF and NiS is 2:1.
Comparing the moles of LiF and NiS, we see that there are 0.347 moles of LiF and 0.143 moles of NiS. Since the ratio between LiF and NiS is 2:1, we can divide the number of moles of LiF by 2 to match the ratio:
0.347 mol LiF / 2 = 0.1735 mol LiF
Now we can compare the calculated number of moles for each compound.
- LiF (0.347 mol) > NiS (0.143 mol)
Based on the calculations, we see that the moles of LiF are greater than the moles of NiS. Therefore, the limiting reactant is NiS, and LiF is the excessive reactant. This means that all 0.143 moles of NiS will react, and some of the LiF will be left over.
In summary:
- Limiting reactant: Nickel (II) sulfide (NiS)
- Excessive reactant: Lithium fluoride (LiF)
Note: The calculations and conclusions may vary if the balanced chemical equation is different.