Answer:
= 110 g AgCl
Step-by-step explanation:
NaCl + Ag(NO3) → AgCl + NaNO3
↓ ↑
mol → mol
NaCl AgCl
[tex]\frac{45 g NaCl }{} \frac{1 mol NaCl}{57.44 g NaCl} \frac{1 mol AgCl}{1 mol NaCl } \frac{143.87 g AgCl}{1 mol AgCl} == 110 g AgCl[/tex]
