1.The reaction of nickel (II) chloride solid with sodium hydroxide aqueous solution occurs according to the balanced reaction equation given below.
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?
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.
To determine the limiting reactant, you need to compare the number of moles of each reactant available.
First, let's convert the given values into moles:
1) Calculate the moles of NiCl2:
Molar mass of NiCl2 = 58.69 g/mol
Moles of NiCl2 = mass / molar mass = 28.37 g / 58.69 g/mol
2) Calculate the moles of NaOH:
Molarity (M) = moles / volume (L)
Convert the volume from mL to L: 400 mL = 0.400 L
Moles of NaOH = Molarity × volume
= 1.85 mol/L × 0.400 L
Now that you have the moles of each reactant, you can compare them to determine the limiting reactant:
Moles of NiCl2 = (28.37 g / 58.69 g/mol) = ...
Moles of NaOH = (1.85 mol/L × 0.400 L) = ...
Compare the coefficients in the balanced equation to determine the stoichiometry. For every 1 mole of NiCl2, 2 moles of NaOH are required.
For the reaction to occur, you need twice the number of moles of NaOH compared to NiCl2. If the moles of NaOH are less than two times the moles of NiCl2, then NiCl2 is the limiting reactant.
If the moles of NaOH are more than two times the moles of NiCl2, then NaOH is the limiting reactant.
Once you determine the limiting reactant, you can calculate the grams of Ni(OH)2 formed from each reactant:
1) If NiCl2 is the limiting reactant:
Use the mole ratio from the balanced equation: 1 mole Ni(OH)2 : 1 mole NiCl2
Moles of Ni(OH)2 = Moles of NiCl2
Then, convert moles of Ni(OH)2 to grams:
Molar mass of Ni(OH)2 = 92.71 g/mol
Mass of Ni(OH)2 = Moles of Ni(OH)2 × Molar mass of Ni(OH)2
2) If NaOH is the limiting reactant:
Use the mole ratio from the balanced equation: 1 mole Ni(OH)2 : 2 moles NaOH
Moles of Ni(OH)2 = (Moles of NaOH) / 2
Then, convert moles of Ni(OH)2 to grams:
Mass of Ni(OH)2 = Moles of Ni(OH)2 × Molar mass of Ni(OH)2
To find the mass of the excess reactant, subtract the moles of the limiting reactant consumed from the initial moles of the excess reactant.
Finally, calculate how much of the excess reactant will be left over by multiplying the moles of the excess reactant by its molar mass.
Let me know if you need help with any of the specific calculations!
To solve this problem, you can start by converting the given quantities into moles. Here are the steps to follow:
1. Calculate the moles of NiCl2:
Using the molar mass of NiCl2 (the sum of the atomic masses of nickel and chloride), you can convert the mass of NiCl2 into moles.
2. Calculate the moles of NaOH:
Given the concentration of the NaOH solution (1.85 M) and its volume (400. mL), you can calculate the moles of NaOH by multiplying the volume (in liters) by the concentration (in moles per liter).
3. Determine the limiting reactant:
Compare the moles of NiCl2 and NaOH to determine which one is the limiting reactant. The reactant that produces fewer moles of product is the limiting reactant.
4. Calculate the moles of Ni(OH)2 produced by each reactant:
Using the balanced equation, you can determine the stoichiometric ratio between NiCl2 and Ni(OH)2. This ratio tells you how many moles of Ni(OH)2 are produced per mole of NiCl2 reacted. Multiply the moles of NiCl2 by this ratio to find the moles of Ni(OH)2 produced by NiCl2. Repeat the same step for NaOH.
5. Convert the moles of Ni(OH)2 to grams:
Using the molar mass of Ni(OH)2, calculate the mass of Ni(OH)2 produced by each reactant.
To find the mass of the excess reactant used in the reaction, subtract the moles of the limiting reactant from the initial moles of that reactant, and use this difference to calculate the mass of the excess reactant. Finally, you can subtract the mass of the excess reactant from the initial mass to determine how much is left over at the end.
Remember to follow the units throughout the calculations and apply the appropriate conversion factors.