Gold can be dissolved from rock ore by treating the rock with sodium cyanide in

Question

Gold can be dissolved from rock ore by treating the rock with sodium cyanide in the presence of oxygen in air:

4Au(s) + 8NaCN(aq) + O2(g) + 2H2O(l) --> 4NaAu(CN)2(aq) + 4NaOH(aq)

First, you want to see what percentage of gold is in the ore. You add sodium cyanide to a 3.027 g rock until all the rock is dissolved. Then you titrate the solution with 2.82 mL of 0.0100 M HCl(aq). What is the percentage of gold in the ore?

What I did was assume that there is 2.82 mL of 0.0100 M NaOH since it was titrated with that amount of HCl, I then used that to calculate the mols of NaOH and used the balanced equation to find grams of Au, then divided the mass of Au by the mass of the rock to get the percentage, but I feel like that is wrong. Any suggestions?

Explanation / Answer

moles of HCl = molarity x volume

= 0.01 x 2.82 / 1000

= 2.82 x 10^-5

moles of HCl = moles of NaOH becuase it is 1:1 titration

from balanced equation :

4 moles of Au --------------------------> 4 moles of NaOH

x moles of Au -----------------------> 2.82 x 10^-5 moles of NaOH

moles of Au = 2.82 x 10^-5

molar mass of gold = 197 g/mol

mass of gold = moles x molar mass

= 2.82 x 10^-5 x 197

= 5.56 x 10^-3 g

= 0.00556 g

percent of gold in ore = gold mass x 100 / ore mass

= 0.00556 x 100 / 3.027

= 0.184 %

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