Question

The zinc content of a 1.23 g ore sample was determined by dissolving the ore in HCI, which reacts with the zinc, and then neutralizing excess HCl with NaOH. The reaction of HCI with Zn is shown below. Zn(s)-2HCl(aq) → ZnCl2(aq)-H,(g) The ore was dissolved in 150 mL of 0.600 M HCI, and the resulting solution was diluted to a total volume of 300 mL. A 20.0 mL aliquot of the final solution required 9.19 mL of 0.529 M NaOH for the HCI present to be neutralized. What is the mass percentage (%w/w) of Zn in the ore sample? Number %Zn

Answer 1

Balanced chemical equation: HCI (aq) + NaOH (aq)Na (aq) + H2O() IL 1000 mL 0.529 mol NaOH 1L No of moles of NaOH 9, 19 mL × 0.00486 mol NaOH 1 mol HC 1 mol NaOH No of moles of HClneeded 0.00486 mol NaOHx 0.00486 mol HCl 1 L Volume og HCI = 20.00 mL×1000 = 0.02000 L 0.00486 m ol HCl 0.02000 mI Concentration of HCl = = 0.243 mol/L 0.243 M Equation: 2HCl + Zn > ZnClh + H2 0.243 mol HCl 1L 1 L No of moles of HCl = 20.00 mL ×--× 1000 mL = 0.00486 mol HCl 1 mol Zn No of moles of Zn needed 0.00486 mol HC1x = 0.00243 mol Zn 65.38 g 1 mol Zn Amount of Zn needed 0.00243 mol Zn x - 0.159 g 0.159&100% o Z1.23 g 13%