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
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 = ??posted by DrBob222