For the following reaction, 5.05 grams of copper are mixed with excess silver nitrate. The reaction yields 11.0 grams of copper(II) nitrate. silver nitrate (aq) copper (s) copper(II) nitrate (aq) silver (s) What is the ideal yield of copper(II) nitrate

14.9 g is the ideal yield of Cu(NO₃)₂

Explanation:

Reactants for the reaction: Cu and AgNO₃

Products: Copper nitrate and Ag

The balanced reaction is: Cu(s) + 2AgNO₃(aq) →  2Ag (s) + Cu(NO₃)₂

As the silver nitrate is in excess, the Cu will be the limiting reagent.

We convert the mass to moles → 5.05 g . 1 mol/ 63.55 g = 0.0794 moles

Ratio is 1:1, so 0.0794 moles will produce 0.0794 moles of Cupper(II) nitrate. We convert the moles to mass, and that value will be the theoretical yield.

0.0794 mol . 187.55 g /1 mol =  of Cu(NO₃)₂

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3