The zinc content of a 1.56 g ore sample was determined by dissolving the ore in

Question

The zinc content of a 1.56 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) --> ZnCl2(aq) + H2(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 9.46 mL of 0.540 M NaOH for the HCl present to be neutralized. What is the mass percentage (%w/w) of Zn in the ore sample?

________ % Zn

Explanation / Answer

Zn(s) + 2HCl(aq) ------------> ZnCl2(aq) + H2(g)

HCl(aq) + NaOH(aq)   -----------> NaCl + H2O(l)

Now, HCl & NaOH reacts in the molar ratio of 1:1

Thus, moles of NaOH = molarity of NaOH solution*volume in litres = 0.54*0.00946 = 0.0051084

Now, moles of excess HCl in the 300 l sample of aliquot = moles of NaOH reacting with 20 ml aliquot*!5

Hence moles of excess HCl = moles of NaOH*15 = 0.0051804*15 = 0.077706

Now, total moles of HCl present = molarity of HCl solution *volume in litres = 0.6*0.15 = 0.09

Thus, moles of HCl reacting with zinc = 0.09 - 0.077706 = 0.012294

Now, HCl & Zn reacts in the molar ratio of 2:1

Thus, moles of Zinc present in the ore = (1/2)*moles of HCl reacting with the ore = 0.006147

Now, molar mass of zinc = 65.38 g/mole

Thus, mass of zinc present in the ore = moles of zinc*molar mass olf zinc = 0.006147*65.38 = 0.402 g

Thus, mass % of Zinc in the ore = (mass of zinc/mass of ore)*100 = (0.402/1.56)*100 = 25.762 %

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.