Question

If the NaOH is added to 35.0 mL of 0.167 M Cu(NO3)2 and the precipitate isolated by filtration, what is the theoretical yield of Cu(OH)2? Cu(NO3)2 (aq) + 2 NaOH (aq) → Cu(OH)2 (s) + 2 NaNO3 (aq)

Answer 1

volume , V = 35.0 mL= 3.5*10^-2 L

we have below equation to be used:

number of mol of Cu(OH)2,

n = Molarity * Volume

= 0.167*0.035

= 5.845*10^-3 mol

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

= 1*63.55 + 2*16.0 + 2*1.008

= 97.566 g/mol

From balanced chemical reaction, we see that

when 1 mol of Cu(NO3)2 reacts, 1 mol of Cu(OH)2 is formed

mol of Cu(OH)2 formed = (1/1)* moles of Cu(NO3)2

= (1/1)*5.845*10^-3 mol

= 5.845*10^-3 mol

we have below equation to be used:

mass of Cu(OH)2 = number of mol * molar mass

= 5.845*10^-3*97.57

= 0.570 g

Answer: 0.570 g