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Answer to Question #56055 in General Chemistry for sue

Question #56055
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.
Expert's answer
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

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