Answer to Question #56055 in General Chemistry for sue

Question

How many grams of Ni(OH)2 are produced from the reaction of 51.0 mL of a 2.00 M NaOH solution and excess NiCl2?
Express your answer with the appropriate units.

Answer 1

Solution:
The chemical reaction of the precipitate formation is:
NiCl₂ + 2NaOH = Ni(OH)₂ + 2NaCl
Then calculate the mass of the precipitate:
n(Ni(OH)₂)=0.5*n(NaOH)=0.5*C(NaOH)*V(NaOH)=0.5*0.05*2=0.05 mol
m(Ni(OH)₂)=n(Ni(OH)2)*M(Ni(OH)₂)=0.05*93=4.65 g

Answer: m(Ni(OH)₂)= 4.65 g