Question

How many grams of Cu(OH)2 will precipitate when excess KOH solution is added to 76.0 mL of 0.764 M Cu(NO3)2 solution

Answer 1

5.66g of Cu(OH)2 can be produced

Step-by-step explanation:

Cu(NO3)2 reacts with KOH, as follows:

Cu(NO3)2 + 2KOH → Cu(OH)2 + 2KNO3

76.0mL of 0.764M Cu(NO3)2 are, in moles:

0.0760L * (0.764mol / L) = 0.0581 moles Cu(NO3)2

As 1 mole of Cu(NO3)2 produce 1 mole of Cu(OH)2, the moles of Cu(OH)2 produced are 0.0581 moles Cu(OH)2.

Molar mass of Cu(OH)2 is 97.561g/mol. In 0.0581 moles there are:

0.0581 moles Cu(OH) * (97.561g / mol) =

5.66g of Cu(OH)2 can be produced