How many grams of copper metal are produced from the reaction of 21.84 ml of 0.108 m cu2+ with excess zinc metal?

The grams  of copper  metal that  are  produced  from  the reaction  of  21.84  ml   of 0.108  M cu2+   is 0.150  grams

calculation

Step  1 :   write the equation for  reaction

Cu^2+(aq)  + Zn(s)→    Cu (s)  +  Zn^ 2+ (aq)

Step 2: calculate  the  moles  of   CU 2+

moles  =   molarity x volume  in liters

molarity = 0.108M=  0.108 mol /l

  volume in liters  = 21.84 /1000 =0.02184  L

moles is therefore= 0.108  mol/l  x 0.02184 L  =0.00236  moles

Step 3 use  the  mole ratio  to determine  moles of Cu

that is CU 2+ : Cu  is 1:1 therefore the  moles of CU  is also =0.00236  moles

step 4:  calculate the  mass of CU

Mass  =  moles x  molar  mass

from  periodic table  the  molar  mass of  Cu  = 63.5 g /mol

mass= 0.00236  moles  x 63.5 g/mol = 0.150  grams

it is stable as it is in group 0 which have full outer shells of electrons