Question

A 1.91-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 0.44 g. What is the percentage by mass of NaNO3 in the original mixture?

I know I have to write the equation and balance it then find the moles of Ag2S (since NO3 does not precipitate), but I'm not sure where to go from there.

Cheers
Right so far. Find mols Ag2S, convert to mols NaNO3, convert to g NaNO3 and stick in the % formula.
%NaNO3 = (mass NaNO3/mass sample)*100 = ?
How do I convert mols Ag2S to mols NaNO3?
bobpursley
wes, use the balanced equation and the coefficients of Ag2S and NaNO2.

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