Question

How many grams of Cu(OH)2 will precipitate when excess Ba(OH)2 solution is added to 77.0 mL of 0.678 M Cu(NO3)2 solution?

Answer 1

Mole ratio for the reaction$=1:1$

$\therefore 1mol Cu(OH)_2 reacts with 1 mol Cu(NO_3)_2 in excess Ba(OH)_2$ Mol=Molarity * Volume (in L)

$=0.678m/L \times 0.077L=0.0522mol$

Molar mass of $Cu(OH)_2=97.561g/mol$

$\therefore 5.12 g of Cu(OH)_2$ Will precipitate