A 1.27-g sample of a mixture of AgNO3 and NaNO3 is treated with excess Na2S(aq). The precipitate is filtered off, dried and weighed. The dried precipitate weighs 0.46 g. What is the percentage by mass of NaNO3 in the original mixture?

Answer 1

The Na2S reacts with AgNO3 but not with NaNO3.
Na2S + 2AgNO3 ==> Ag2S + 2NaNO3
mols Ag2S = 0.46/molar mass Ag2S = ?
Using the coefficients in the balanced equation, convert mols Ag2S to mols AgNO3.
Since the mixture has a mass of 1.27 g, 1.27- mass AgNO3 from above = mass NaNO3.
Then % NaNO3 = (mass NaNO3/mass sample)*100 = ?