Răspuns:
3,06g 1,17g x g
AgNO₃ + NaCl --> AgCl↓ + NaNO₃
170g 58,5g 143,5g
3,06/160=0,018
iar 1,17/58,5=0,02 => NaCl in exces cu m exces= 0,02-0,019=0,002 g
Calculam masa de AgCl (pp) dupa AgNO₃ => x- 3,06*143,5/170=2,583 g pp
Calculam masa de substanta pura( de pe reactia chimica) de NaCl care s-ar obtine din 3,06 g AgNO₃.
3,06 g y
AgNO₃ + NaCl --> AgCl↓ + NaNO₃ ==> y= 3,06*58,5/170=1,053 g NaCl pur
170 g 58,5 g
p/100 = m pur/ m imp => p/100=1,053/1,17 => p/100=0,9 => p%=90%
