Question

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

A.) How many milliliters of 0.200M NaOH solution are needed to react with 48.0 mL of a 0.240 M NiCl2 solution? Express your answer with the appropriate units.

B.) How many grams of Ni(OH)2 are produced from the reaction of 34.0 mL of a 1.70 M NaOH solution and excess NiCl2? Express your answer with the appropriate units.

C.) What is the molarity of 30.0 mL of a NiCl2 solution that reacts completely with 12.2 mL of a 0.250 M NaOH solution? Express your answer with the appropriate units.

Answer 1

A)

Here:

M(NiCl2)=0.24 M

M(NaOH)=0.2 M

V(NiCl2)=48.0 mL

According to balanced reaction:

2*number of mol of NiCl2 =1*number of mol of NaOH

2*M(NiCl2)*V(NiCl2) =1*M(NaOH)*V(NaOH)

2*0.24 M *48.0 mL = 1*0.2M *V(NaOH)

V(NaOH) = 115.2 mL

Answer: 115 mL

B)

lets calculate the mol of NaOH

volume , V = 34 mL

= 3.4*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 1.7*3.4*10^-2

= 5.78*10^-2 mol

According to balanced equation

mol of Ni(OH)2 produced = (1/2)* moles of NaOH

= (1/2)*5.78*10^-2

= 2.89*10^-2 mol

This is number of moles of Ni(OH)2

Molar mass of Ni(OH)2,

MM = 1*MM(Ni) + 2*MM(O) + 2*MM(H)

= 1*58.69 + 2*16.0 + 2*1.008

= 92.706 g/mol

use:

mass of Ni(OH)2,

m = number of mol * molar mass

= 2.89*10^-2 mol * 92.71 g/mol

= 2.679 g

Answer: 2.68 g

C)

Here:

M(NaOH)=0.25 M

V(NaOH)=12.2 mL

V(NiCl2)=30.0 mL

According to balanced reaction:

1*number of mol of NaOH =2*number of mol of NiCl2

1*M(NaOH)*V(NaOH) =2*M(NiCl2)*V(NiCl2)

1*0.25*12.2 = 2*M(NiCl2)*30.0

M(NiCl2) = 0.0508 M

Answer: 0.0508 M