a 6.54 g sample consisting of a mixture of silver nitrate and sodium nitrate is dissolved in water. this mixture then reacts with barium chloride to for 3.50 g of silver chloride. calculate the percent by mass of silver nitrate in the first mixture

Answer 1

2AgNO3 + BaCl2 ==> 2AgCl + Ba(NO3)2

How many moles AgCl were formed? That is g/molar mass = moles = 3.50/143.32 = 0.0244 moles.

Since 1 mole AgCl = 1 mole AgNO3, we must have had 0.0244 moles AgNO3 in the original sample and we convert to grams by g = moles x molar mass.

Then percent AgNO3 = (mass AgNO3/mass sample)*100 = ??

Answer 2

%AgNO3 is 10.38%