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

The balanced chemical equation is as follows:
Ni(s) + 2AgNO3(aq) ==> Ni(NO3)2(aq) + 2Ag(s)
Let X = amount of Ni that reacted.
Then X reacts to form Ag. How many grams Ag are formed. That's
2(X)*107.87/58.69
If X = amount of Ni that reacted, how many grams Ni are left. that's
21.5 g - X.
grams Ag formed + grams Ni left = 36.5 grams. Put all of that together.
21.5 - X + 2(X)*107.87/58.69 = 36.5 and solve for X.
I get X = about 5.6 g Ni that reacted. So 21.5 - 5.6 = 15.9 grams Ni left. So 36.5 total metal - 15.9 g Ni remaining = 20.6 grams Ag formed.
I checked it by asking how much Ag would be formed from the 5.6 g Ni that reacted. That's 5.60 x 2*107.87*X/58.69 = 20.59 grams Ag formed.
Interesting problem.