a, AgNO3 + NaCl ---> NaNO3 + AgCl↓
nNaCl=\(\dfrac{11,7}{58,5}=0,2mol\)
nAgNO3=\(\dfrac{25,5}{170}=0,15mol\)
=> NaCl dư
mAgCl↓=143,5.0,15=21,525 g
b, nNaCl dư=0,2-0,15=0,05 mol
nNaNO3=nAgNO3=0,15 mol
=> ndd=0,05+0,15=0,2 mol
CM NaCl dư = \(\dfrac{0,05}{0,2}=0,25M\)
CM NaNO3 = \(\dfrac{0,15}{0,2}=0,75M\)