1.The reaction of nickel (II) chloride solid with sodium hydroxide aqueous solution occurs according to the balanced reaction equation given below.

Question

NiCl2 (s) + 2 NaOH(aq) → Ni(OH)2(s) + 2 NaCl(aq)

If 28.37 g of solid NiCl2 are mixed with 400. mL of 1.85 M NaOH solution and allowed to react, determine which reactant is limiting. Show two calculations to g of Ni(OH)2 formed by each reactant.

Calculate the mass of excess reactant in grams that will be used up in the reaction, and use it to determine how much will be left over at the end.

So, I am looking on guidance on how to go forward with this problem if you can offer it. I asked my professor about it, but I usually get faster and better replies on here.

Where do I start? For the first part, I want to know if there is a certain formula? Do i convert all my stuff into moles?

Answer 1

There are two ways to do limiting reagent (LR) problems. The long way and the short way. Yes, you convert everything to moles for either approach.
.....................NiCl2(s) + 2NaOH(aq) → Ni(OH)2(s) + 2NaCl(aq)
long way first: I've estimated for speed but you need to take that into consideration. I've not been careful with significant figures so watch that.
mols NiCl2 = g/molar mass = 28.37/130 = about 0.22
mols NaOH =M x L = 1.85 x 0.400 = 0.74
Now do two stoichiometry problems, first with NiCl2 then with NaOH.
How much Ni(OH)2 from NiCl2 and excess NaOH. That's
0.22 mols NiCl2 x(1 mol Ni(OH)2/1 mol NiCl2) = 0.22 mols Ni(OH)2. That times molar mass gives grams with one calculation.
Next do the same with the NaOH. That's
0.74 mols NaOH x (1 mol Ni(OH)2/2 mol NaOH) = 0.37 mols NiCl2. That times molar mass gives grams which is a second calculation.
In LR problems the small number always wins so you will form 0.22 mols Ni(OH)2, which is the LR with NaOH being the excess reagent (ER). Apparently the problem want you to choose between the two smaller amount of grams but I never do the grams because moles gives you the same answer with LR and ER.
So g Ni(OH)2 = mols x molar mass = ?
The short way of determining the LR and ER is next.
You have 0.22 mol NiCl2. It needs 0.44 mols NaOH and you have that much and more so NiCl2 is the LR and NaOH is the ER.
To calculate ER, how much ER is used up when the 0.22 mols NiCl2 reacts.
That's 0.22 mols NiCl2 x (2 mol NaOH/1 mol NiCl2) =0.44 mols. You had 0.74 initially. The excess is 0.74-0.44 = 0.30
grams NaOH = 0.30 mols x 40 g/mol = 12 g not used. You had 0.74 x 40 = 30 g initially so you have 30-12 = ? grams NaOH not used.
Hope this is helpful. Since this is below the end of the page I don't know that you will that so I'm posting this at the top also. Check my work. It's late and I make a lot of typos.