Question

The zinc content of a 1.45 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+2HCL--ZnCl2+H2

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 8.48 mL of 0.508 M NaOH for the HCl present to be neutralized. What is the mass percentage (%w/w) of Zn in the ore sample?

Answer 1

Zn + 2 HCl => ZnCl2 + H2 HCl NaOH> NaCH20 Excess moles of HC in 20 mL of diluted solutiorn moles of NaOH-volume x concentration of NaOH 8.48/1000 x 0.508 0.00430784 mol Excess moles of HCl in 300 mL of diluted solution 300/20 x 0.00430784 0.0646176 mol Initial moles of HCl initial volume x initial concentration of HCl #150/1000 x 0.600 0.09 mol Moles of reacted HCl 0.09 0.0646176 0.0253824 mol Moles of Zn1/2 x moles of reacted HCl #1 /2 x 0.0253824 0.0126912 mol Mass of Zn moles x molar mass of Zn 0.0126912 x 65.380.8298 g Percent of Zn-mass of Zn/mass of sample x 100% 0.8298/1.45 x 100%-57.2%