Question

How many grams of Ni(OH)2 are produced from the reaction of 30.5 mL of 1.87 M NaOH? NiCl2(aq)+2NaOH(aq)→Ni(OH)2(s)+2NaCl(aq)

Answer 1

we have the Balanced chemical equation as:

NiCl2(aq)+2NaOH(aq)→Ni(OH)2(s)+2NaCl(aq)

lets calculate the mol of NaOH

volume , V = 30.5 mL

= 3.05*10^-2 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 1.87*0.0305

= 5.704*10^-2 mol

From balanced chemical reaction, we see that

when 2 mol of NaOH reacts, 1 mol of Ni(OH)2 forms

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

= (1/2)*5.704*10^-2

= 2.852*10^-2 mol

This is number of moles of Ni(OH)2

Molar mass of Ni(OH)2 = 1*MM(Ni) + 2*MM(O) + 2*MM(H)

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

= 92.706 g/mol

we have below equation to be used:

mass of Ni(OH)2,

m = number of mol * molar mass

= 2.852*10^-2 mol * 92.706 g/mol

= 2.64 g

Answer: 2.64 g