Question

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

1. How many milliliters of 0.165 M NaOH are needed to react with 21.9 mL of 0.465 MNiCl2?

2. How many grams of Ni(OH)2 are produced from the reaction of 34.9 mL of 1.87 MNaOH?

Answer 1

In the given reaction,

NiCl2 + 2NaOH --> Ni(OH)2 + 2NaCl

1 mol of NiCl2 reacts with 2 mols of NaOH

1. mols of NiCl2 present = 0.465 M x 21.9 ml = 10.1835 mmol

mols of NaOH neede = 2 x 10.1835 = 20.367 mmol

Volume of NaOH needed = 20.367 mmol/0.165 M = 123.44 ml

2. mols of NaOH used = 1.87 M x 34.9 ml = 65.263 mmol

mols of Ni(OH)2 produced = 65.263 mmol/2 = 32.6315 mmol

grams of Ni(OH)2 produced = 32.6315 mmol x 92.708 g/mol/1000 = 3.025 g

[Formulas used, molarity = moles/L of solution ; moles = grams/molar mass]