i cant do this chemistry problem ive already tried 3 times....?

Question

To extract gold from its ore, the ore is treated with sodium cyanide solution in the presence of oxygen and water.
4 Au(s) + 8 NaCN(aq) + O2(g) + 2 H2O(l) 4 NaAu(CN)2(aq) + 4 NaOH(aq)

(a) Determine the mass of gold that can be extracted if 24.9 g sodium cyanide is used.
g
(b) If the mass of the ore from which the gold was extracted is 167.5 g, what percentage of the ore is gold?
%

a) i know for a fact is 50.05 g.... i just need b and or how to do it

Answer 1

(a)
Use stoichiometry:

Moles of NaCN (sodium cyanide) = mass present/molecular mass
n (moles) NaCN = 24.9g/49.0072g/mol
n NaCN = 0.5081 mol

NOW:
for every 8 moles of NaCN you have 4 moles of gold produced. Therefore, the number of moles of gold is half the number of moles of NaCN

So--> moles Au = 0.5 x n (NaCN)
n Au = 0.5 x 0.5081 moles
n Au = 0.254 moles

We know that the molar mass(the molecular weight) of gold is 196.966569g /mol, so you find the mass by mulitplying number of moles by molecular weight:

mass Gold(Au) = n (Au) x Molecular weight
mass Gold = 0.254 g x 196.966569 g/mol
mass gold = 50 g

(b)
% = (mass of substance of interest/total mass) x 100
% = (50g/167.5g) x 100
% = 29.85 %

BF
4th yr Chem Major
U of O

Answer 2

The equation indicates that for each 2 moles of NaCN, 1 g-atom of gold is dissolved.

a) convert 24.9 g of NaCN to moles (about 0.5 mole), which indicates about 0.25 g-atoms of gold is dissolved. Gold has a atomic weight of about 200, so the mass should be about 50 grams.

b) per-cent purity = 100 (mass of gold recovered)/
mass of ore treated.
Just plug it in

Answer 3

percentage of gold in ore = (mass of gold/mass of ore)(100)
= (50.05 / 167.5)(100)

Answer 4

Haven't ever thought about it in that way to be honest