How many grams of Cu(OH)2 will precipitate

Question

How many grams of Cu(OH)2 will precipitate when excess KOH solution is added to 61.0 mL of 0.521 M Cu(NO3)2 solution?

 

Cu(NO3)2(aq) + 2KOH(aq) Cu(OH)2(s) + 2KNO3(aq)

 

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Answer 1

             Cu(NO3)2(aq)       +      2KOH(aq)   ------------    Cu(OH)2(s)    +    2KNO3(aq)

 

If you look at the coefficient:

 

 

1 mol Cu(OH)2 will precipitate when 1 mol Cu(NO3)2 reacts with excess KOH.

 

 

The mol of Cu(NO3)2 added = 0.521 mol/L x 0.061 L = 0.032 mol (FYI: mol  = Molarity X Volume(in L))

 

 

Thus,  [1 mol Cu(OH)2/1 mol Cu(NO3)2] x 0.032 mol Cu(NO3)2 = 0.032 mol Cu(OH)2 will precipitate.

 

 

Find the molar mass of Cu(OH)2, and convert 0.032 mol to gram.

 

 

FYI: 3.12 g of Cu(OH)2 will precipitate.

 

 

Answer 2

Hello Hussein

 

Since you are given volume and Molarity of Cu(NO₃)₂, you can calculate the # of moles.

 

mol  =  Molarity x Volume  = 0.521 x 0.0610 L = 0.0318 mol Cu(NO₃)₂

 

From the chemical reaction, you can see that for every 1 mole of Cu(NO₃)₂ that reacts, 1 mole of Cu(OH)₂ is formed.

 

0.0318 mol Cu(NO₃)₂  x  1 mol Cu(OH)₂ / 1 mol Cu(NO₃)₂  =  0.0318 mol Cu(OH)₂

 

Convert mol Cu(OH)₂ to grams using its molar mass (98 g/mol)

 

0.0318 mol Cu(OH)₂  x  98 g Cu(OH)₂ / 1 mol Cu(OH)₂  =  3.12 g Cu(OH)₂