设原混合物中含NaClxmol,含KIymol,则：
58.5x + 166y = 7.81……………………………………………………………（1）
58.5x + 74.5y = 4.15……………………………………………………………（2）
解（1）和（2）得：
x =0.02 y = 0.04
(1)m(NaCl) = 58.8g/mol * 0.02mol = 1.17g
w(NaCl) = 1.17g/7.81g*100% = 15.0%
(2) n(AgCl) = n(NaCl) = 0.02mol
m(AgCl) = 143.5g/mol * 0.02mol = 2.87g
n(AgI) = n(KI) = 0.04mol
m(AgI) = 235g/mol * 0.04mol = 9.4g
沉淀总质量 ＝ 87g + 9.4g = 12.27g
答：……………………………………………………