Question

A mixture of NaBr, Nal and NaNO3 weights 0.6500 g. By treating it with AgNO₃ a halide precipitated is formed weighing 0.9390 g. By heating this precipitate in Cl₂ current it becomes AgCl that weights 0.6566 g. What is the percentage of NaNO₃ in the original sample?

Answer 1

No of mol of AgCl = 0.6566/143.32 = 0.00458 mol

No of mol of Cl = 0.00458

total no of mol of Br+I = 0.00458 mol

mole ratio of I,Br = 79.904/126.90447 = 0.63

No of mol of Br = 0.00458/1.63 = 0.0028 mol

no of mol of I = 0.00458 - 0.0028 = 0.00178 mol

mass of NaBr = 0.0028*102.894 = 0.288 grams

mass of NaI = 149.89*0.00178 = 0.267 grams

totla mass of NaBr+ NaI = 0.288+0.267 = 0.555 grams

mass of NaNO3 = 0.65-0.555 = 0.095 grams

percent = 0.095/0.65*100 = 14.61%