I need help with this chemistry question on moles?

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Chemchick

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"A silver coin was analyzed and found to contain only silver and copper. 1.580 g of this coin was dissolved in concentrated nitric acid and the resultant solution diluted. reaction of the solution with excess hydrochloric acid yielded 1.050 g of silver chloride. calculate the percentage of silver in the sample
 
Mr AgCl = 108+35.5 = 143.5

1.050 g of silver chloride = 1.050 g AgCl = 1.05/143.5 = 3/410 moles of AgCl

mol of Ag = mol of Cl = mol og AgCl = 3/410 mol

We know that the total mol of AgCl is constant during both reaction,
so total mass of Ag = 3/410 * 108 = 162/205 gram

percentage of silver = total mass of Ag/total mass of the coin * 100%
% Ag = (162/205)/1.580 * 100%
% Ag =50.015 %
 
We got 1.050 gms of AgCl so, moles of AgCl = moles of Ag
Moles of Ag = 1.050/(108 + 35.5) = 0.00731

So, weight of Ag = 0.00731 * 108 = 0.7498 gms
So, % of silver = Mass of silver/(Total Mass)*100 = 49.967 = 50%
 
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