Question

The zinc content of a 1.75 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.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 9.04 mL of 0.536 M NaOH for the HCl present to be neutralized. What is the mass percentage (%w/w) of Zn in the ore sample?

Answer 1

Ans :

Number of mol of HCl = molarity x volume (L)

= 0.600 x 0.150

= 0.09 mol

The solution was then diluted to 300 mL , mol of HCl remained 0.09 mol in it

0.09 mol is present in 300 mL

So number of mol in 20 mL = (20 x 0.09) / 300 = 0.006 mol

Number of mol of NaOH = 0.536 x 0.00904

= 0.00484 mol

Each mol NaOH utilises 1 mol HCl according to the reaction

NaOH + HCl = NaCl + H2O

So number of mol of HCl unutilised (or that reacted with Zn) = 0.006 - 0.00484 = 0.00115 mol

0.00115 mol HCl reacted in 20 mL solution

So number of mol that reacted in 300 mL will be : ( 300 x 0.00115) / 20

= 0.0173 mol

Zn reacts with HCl as :

Zn + 2HCl = ZnCl2 + H2

So number of mol of Zn = 0.0173 / 2 = 0.0086 mol

Mass of Zn = mol x molar mass

= 0.0086 x 65.38

= 0.566 g

%w/w of zinc in ore = ( mass of zinc / mass of ore) x 100

= ( 0.566 / 1.75) x 100

= 32.35 %