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