The zinc content of a 1.72 g ore sample was determined by dissolving the ore in
ID: 916130 • Letter: T
Question
The zinc content of a 1.72 g ore sample was determined by dissolving the ore in HCI, which reacts with the zinc, and then neutralizing excess HCI with NaOH. The reaction of HCI with Zn is shown below. Zn(s) + 2HCl(aq) Right arrow ZnCI2(aq)+H2(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 9.95 mL of 0.550 M NaOH for the HCI present to be neutralized. What is the mass percentage (%w/w) of Zn in the ore sample?Explanation / Answer
From M1V1= M2V2
Molarity of HCl in 20ml for neutralization of NaOH
20*M1= 9.95*0.55
M1= 0.273625 M
Moles of HCl in 300 ml =0.273625*300/1000= 0.0821 Moles
Initial moles of HCl = 150*0.6/1000= 0.09
Moles of HCL consumed by Zn+2HCl-à ZnCl2+H2 is 0.09-0.0821=0.0079 moles
From the reaction 2 moles of HCl require 1 mole of Zn
0.0079 moles of HCl require =0.0079/2 =0.00395 moles of Zn
Mass of Zinc = atomic weight* moles of Zinc = 64*0.00395=0.2528gms
Initial mass of sample =1.72
Mass % of Zinc in the sample =100*( 0.2528/1.72)=14.7%
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