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

User Defnull
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1 Answer

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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

User RemusS
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