Ag + 2HNO3 → AgNO3 + NO2↑ + H2O (1)
AgNO3 + HCl → AgCl↓ + HNO3 (2)
nAgCl =\(\frac{{0,398}}{{143,5}}(mol)\)
Theo PTHH (2) Tính được nAgNO3 = nAgCl = \(\frac{{0,398}}{{143,5}}(mol)\)
Theo PTHH (1) => nAg = nAgNO3= \(\frac{{0,398}}{{143,5}}(mol)\)
=> % Ag = \(\frac{{{m_{Ag}}}}{{m{\,_{hop\,kim}}}}.100\% = \frac{{\frac{{0,398}}{{143,5}}.108}}{{0,5}}.100\% = 59,9\% \approx 60\% \)