Question

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Answer 1

Answer:

= 110 g AgCl

Step-by-step explanation:

NaCl  +  Ag(NO3)  →   AgCl  +  NaNO3

         ↓                               ↑  

        mol           →           mol

      NaCl                         AgCl

$\frac{45 g NaCl }{} \frac{1 mol NaCl}{57.44 g NaCl} \frac{1 mol AgCl}{1 mol NaCl } \frac{143.87 g AgCl}{1 mol AgCl} == 110 g AgCl$

Answer 2

The correct
answer is 110.