x g of Ag was dissolved in nitric acid and the solution was treated with excess of NaCl when 2.87g of AgCl was precipitated.the value of x is
The molecular weight of AgCl is M(agcl)= 143.32 g.mol-1
You have m (AgCl)= 2.87 g of AgCl at the end so that means you have:
m(agcl)/M(AgCl) = 2.87g/143.32g.mol-1 = 0.020 moles of AgCl or n(agcl).
Since we know the reaction was complete, it means you had 0.020 moles of silver in the coin: n(ag)=n(agcl).
M(ag) = 107.868 g.mol-1
So your weight of silver in the coin was:
m(ag)=n(ag)xM(ag)= 0.20 mol x 107.868 g.mol-1
So m(ag) = 21.57 g
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