Question

A 2.31 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 0.47 g. What is the percentage by mass of NaNO3 in the original mixture?

Your answer must be accurate to two significant digits. Keep several extra digits in your calculations and round off only at the end. Do not include the percent sign (%) as part of your answer!

Molar masses (in g/mol):

N, 14.01

O, 16.00

Na, 22.99

S, 32.07

Cl, 35.45

Ag, 107.9

Answer 1

2 AgNO3 (aq)   + Na2S (aq) -----------------> Ag2S (s) + 2 NaNO3 (aq)

mass of precipitate = 0.47 g

moles of Ag2S = 0.47 / 247.8 = 1.897 x 10^-3

2 mol AgNO3 ---------------> 1 mol Ag2S

?? mol AgNO3   -------------> 1.897 x 10^-3 mol Ag2S

moles of AgNO3 = 1.897 x 10^-3 x 2 = 3.793 x 10^-3 mol

moles of AgNO3 = 3.793 x 10^-3 mol

mass of AgNO3 = 3.793 x 10^-3 x 169.87 = 0.644 g

mass of NaNO3 = 2.31 - 0.644 = 1.67 g

mass % of NaNO3 = 1.67 / 2.31) x 100 = 72.1 %

mass % of NaNO3 = 72.1 %