The zinc content of a 1.17 g ore sample was determined by dissolving the ore in

Question

The zinc content of a 1.17 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 8.50 mL of 0.507 M NaOH for the HCl present to be neutralized. What is the mass percentage (%w/w) of Zn in the ore sample?

Explanation / Answer

m = 1.17 g of ore

V = 150 ml of acid

M = 0.6 HCl

total V = 300 ml

V = 20 ml of solution needed

V = 8.5 ml of NaOH

M = 0.507 NaOH

first, calculate base used

mol of base = MV = 8.5*0.507 = 4.3095 mmol of NaOH

then; in those 20 ml of solution, there are

4.3095mmol of NaOH :4.3095 mmol of HCl

find concnetration

HCL = mmol/V = 4.3095 /20 = 0.215475 M of HCl

that is the concnetration in the V = 300 ml sample

find moles in sample

mol = M*V = 300*0.215475 = 64.6425 mmol of HCl were "left" since all other Hcl was reacted

total HCl added in the begginig = MV = 150*0.6 = 90 mmol

then

HCl used in reaction of Zn = 90 - 64.6425 = 25.3575 mmol of HCl was used in the reaction of Zn

then

since ratio is 1:2

Zn mol = 25.3575/2 = 12.67875 mmol of Zn was in the sample

MW Zn = 65.38 g/mol

mass of Zn = mol*MW

mass Zn = (12.67875*10^-3) *(65.38) = 0.828936675 g of Zn

% mass = mass of Zn / mass of ore * 100 = 0.828936675/1.17 *100 = 70.8492 % of Zn in sample

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