Question

How many of Ni(OH)2 are produced from the reaction of 35.0 mL of a 0.200 M NaoH solution? grams There is an excess amount of Nicl2. Nicl2(aq) 2NaoH (aq) Ni (oH)2(s) 2Nacl(aq)

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Answer #1

mol of NaOH = MV = 0.2*35 = 7 mmol of NaOH

2 molo = 1 mol

7 mmol of NaOH = 7/2 = 3.5 mmol of NiCl2

mass = mmol*MW = 3.5*129.5994 = 453.5979 mg = 0.4536 g of NiCl2

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