finding mass when reaction has stopped

Question

A 21.5 g sample of nickel was treated with excess silver nitrate solution to produce silver metal and nickel(II) nitrate. The reaction was stopped before all the nickel reacted, and 36.5 g of solid metal (nickel and silver) is present. Calculate the mass of solid silver metal present in grams.

Answer 1

Ni+2AgNO3→Ni(NO3)2+2Ag

21.5g Ni/58.69g =0.366mol Ni

Ag=107.87

Ni=58.69

But, 2molAg per 1 mol Ni, so, change in mass is more like...

twice as many moles Ag as Ni, and then mass relationship is 107.87gAg/58.69, but due to the 2Ag:1Ni mol ratio, the change in mass is like:

107.87(2)/58.69=3.68 times increase

So, final mass of Ag is 3.68x, and the change in mass of Ni is x

36.5=(21.5-x)+(3.68x)

solve for x

x=5.597

3.68x=g Ag

3.68(5.597)=20.59

21.5-x=g Ni after reaction

21.5-5.597=15.903g Ni

20.59gAg+15.903g Ni=36.5g metal