How many grams of CuOH2 will precipitate when excess KOH sol

How many grams of Cu(OH)₂ will precipitate when excess KOH solution is added to 54.0 mL of 0.656 M Cu(NO₃)₂ solution?

b.  What volume of a 0.121 M nitric acid solution is required to neutralize 13.3 mL of a 0.164 M calcium hydroxide solution?

mL nitric acid

Solution

Ans :

a) Number of moles of Cu(NO3)2 = 0.656 x 0.054 = 0.035424 mol

So the number of moles of Cu(OH)2 will be 0.035424 mol

Then the mass of compound precipitated = 0.035424 x molar mass

= 0.035424 x 97.561

= 3.456 grams

b) The reaction is given as :

2HNO3 + Ca(OH)2 = Ca(NO3)2 + 2H2O

Using the formula , M1 V1 / n1 = M2 V2 / n2

where M is molarity , V is volume and n is number of moles

(0.121 x V ) / 2 = 0.164 x 13.3

V = 36.05 mL nitric acid