Question

How many grams of nickel ii chloride are required to neutralize 75.0 ml of 0.555 m lithium hydroxide?

Answer 1

2.70 g of NiCl₂

Step-by-step explanation:

First, we need to know the reactants. Nickle II chloride is formed by Ni⁺² (the II informs the oxidation) and Cl⁻, so it is NiCl₂. Lithium is from group 1 of the periodic table, so its ion is Li⁺, and the hydroxide has the ion OH⁻, so it is LiOH.

The reaction will be a double replacement and the cations (positive ions) will change between them:

NiCl₂ + 2LiOH → Ni(OH)₂ + 2LiCl

So, it is necessary 1 mol of NiCl₂ to react with 2 moles of LiOH. In 75 mL(0.075L) of LiOH with 0.555 M, will have:

n = 0.555x0.075 = 0.041625 moles.

1 mol of NiCl₂ ----------------- 2 moles of LiOH

x ----------------------------------- 0.041625 moles of LiOH

By a simple direct three rule:

2x = 0.041625

x = 0.02081 moles of NiCl₂.

Nickel has molar mass(M) equal to 58.7 g/mol, and chlorine, 35.5 g/mol, so the molar mass of NiCl₂ is 58.7 + 2x35.5 = 129.7 g/mol. The mass is:

m = Mxn

m = 129.7x0.02081

m = 2.70 g of NiCl₂

Answer 2

The neutralization reaction between nickel II chloride and lithium hydroxide is written below.
Ni(Cl)₂ + 2LiOH -> 2LiCl + Ni(OH)₂
From the double replacement reaction, it can be seen that for every 2 moles of lithium hydroxide, we need a mole of Ni(Cl)₂.
0.555 m LiOH x (1 mole Ni(Cl)₂/2 moles LiOH) = 0.2775 moles Ni(Cl)₂
Then, we multiply this number of moles by the molar mass of the NiCl₂ which is equal to 129.6 g/mol. The answer would be
35.964 grams of NiCl₂