How many grams of Cu(OH) 2 willprecipitate when excess KOH solution is added to

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Chemical reaction : Cu(NO3)2 + 2 KOH----> Cu(OH)2 +2KNO3 Given data Volume of Cu(NO3)2 = 77.0mL                                                      = ( 77.0 mL)(1L/1000mL)                                                      = 0.077 L                    Molarity of Cu(NO3)2 = 0.620 M                                                       = 0.620 moles/L                We know that Molarity = moles/Volume (L)                Moles ofCu(NO3)2       = (0.620 moles/L)(0.077 L)                                                     = 0.04774 moles               we know that mole = Mass/molar mass              from the reaction 1 mole of Cu(NO3)2 give onemole of Cu(OH)2              so we need 0.04774 moles of Cu(OH)2                Mass =( moles)(Molar mass)                              = (0.04774 moles)(97.6 g/mol)    [ Cu(OH)2 molarmass = 97.6 g/mol ]                               = 4.66 g 4.66 grams of Cu(OH)2 willprecipitate when excess KOH solution is added to77.0 mL of 0.620 MCu(NO3)2 solution.

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