Question

The zinc content of a 1.35 g ore sample was determined by dissolving the ore in HCl, which reacts with the zinc, and then neutralizing excess HCl with NaOH. The reaction of HCl with Zn is shown below.

Zn(s) + 2HCl (aq) -> ZnCl_2 (aq) +H_2(g)

The ore was dissolved in 150 mL of 0.600 M HCl, and the resulting solution was diluted to a total volume of 300 mL. A 20.0 mL aliquot of the final solution required 9.28 mL of 0.548 M NaOH for the HCl present to be neutralized. What is the mass percentage (%w/w) of Zn in the ore sample?

Answer 1

1 mole of Zn needs 2 moles of HCl

initial moles HCl = (Molarity HCl)(L HCl) = (0.600)(0.150) = 0.0900 moles HCl

Excess HCl = neutralized by NaOH

HCl + NaOH ---> NaCl + H2O

Moles of HCl = Moles of NaOH = Molarity*V in L = 9.28*0.548/1000 = 5.085 millimoles

Moles of HCl in 20 ml aliquot solution = 5.085 millimoles

In 300 ml solution = 5.085 millimoles *300 / 20 = 0.0763 moles

moles HCl that reacted with ore = initial moles HCl - moles of unreacted HCl = 0.0900 - 0.076275 = 0.013725

Moles of Zn = 0.013725 / 2 = 6.8625*10^-3

Mass = moles *molar mass = 0.4486 gms

% --> 0.4486*100 / 1.35 = 33.2 %