Question

Answer the following for the reaction: NiCl2(aq)+2NaOH(aq)→Ni(OH)2(s)+2NaCl(aq)

1.How many milliliters of 0.200M NaOH solution are needed to react with 26.0 mL of a 0.420 M NiCl2 solution?

2.How many grams of Ni(OH)2 are produced from the reaction of 48.0 mL of a 1.80 M NaOH solution and excess NiCl2 ?

3.What is the molarity of 30.0 mL of a NiCl2 solution that reacts completely with 11.3 mL of a 0.360 M NaOH solution?

Answer 1

the given reaction is

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

from the reaction

we can see that

moles of NaOH required = 2 x moles of NiCl2

also

moles = molarity x volume

so

m x v of NaoH = 2 x m x v of NiCl2

so

0.2 x v = 2 x 0.42 x 26

v = 109.2

so

109.2 ml of NaOH is required

2)

now

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

moles of Ni(OH)2 produced = 0.5 x moles of NaOH

now

moles of NaoH = molairty x volume (L)

moles of NaOH = 1.8 x 48 x 10-3

moles of NaOH = 0.0864

so

moles of Ni(OH)2 produced = 0.5 x 0.0864

moles of Ni(OH)2 produced = 0.0432

now

mass = moles x molar mass

so

mass of Ni(OH)2 produced = 0.0432 x 92.708

mass of Ni(OH)2 produced = 4

so

4 grams of Ni(OH)2 is produced

3)

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

m x v of NaOH = 2 x m x v of NiCl2

so

0.36 x 11.3 = 2 x m x 30

m = 0.0678

so

the molarity of NiCl2 is 0.0678 M