In a procedure developed to determine the percent zinc in post 1982 pennies, 50

Question

In a procedure developed to determine the percent zinc in post 1982 pennies, 50 mL of an HCl solution was used to react (dissolve) all of the zinc in the penny. To ensure complete reaction, the solution contains twice as many moles of HCl that is usually needed. What concentration of HCl should be used? What is the amount in moles of excess (unreacted) HCl in solution? How many moles of NaOH would be needed to completely react with all of the excess HCl?

To determine the percent zinc in the penny, the excess was titrated with NaOH. determine the concentration of NaOH needed if you want to use approx 25 mL of NaOH to titrte the excess HCl.

Explanation / Answer

Solution :-

mass of penny is 2.500 g

from 1982 to present the composition of the penny is 97.5 % Zn and 2.5 % Cu

Therefore the mass of the Zn in the penny = 2.500 g * 97.5 % / 100 % = 2.4375 g

now lets calculate the moles of Zn

moles of Zn = mass / molar mass

                 = 2.4375 g /65.39 g per mol

                 = 0.03728 mol Zn

Reaction equation of the Zn and HCl

Zn + 2HCl ------ > ZnCl2 + H2

now lets calculate the moles of the HCl needed to react with 0.03728 mol Zn

0.03728 mol Zn * 2 mol HCl / 1 mol Zn = 0.07456 mol HCl but since the twice amount of the HCl is used to ensure the complete reaction

therefore the moles of HCl used actual = 0.07456 mol * 2 =0.1491 mol

now lets calculate the molarity of the HCl

molarity of HCl = moles / volume in liter

                      =0.1491 mol / 0.050 L

                      = 2.98 M

As half of the moles of HCl reacts with Zn therefore the concnetration of the HCl remain unreacted = 2.98 M /2 = 1.49 M

Now lets calculate the molarity of the NaOH needed to neutralize the HCl

mole ratio of the HCl and NaOH is 1 : 1

therefore the moles of NaOH needed to react with excess HCl = 0.07456 mol

molarity of the NaOH = volume of HCl * molarity of HCl / volume of NaOH

                              = 50 ml * 1.49 M /25 ml

                              = 2.98 M

So we need 2.98 M 25 ml NaOH to neutralize the excess HCl

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