Answer:
25.40 grams Silver Chloride
Step-by-step explanation:
We will sove this problem by using the stoichiometry based on the balanced equation of the reaction
AgNO3 + NaCl ------------------------ AgCl + NaNo3
we are given the amount in grams of silver nitrate so convert it to mol and then calculate mol silver chloride produced and convert this quantiy to grams which is what is asked.
MW AgNO3 : 169.87 g/mol
MW AgCl: 143.32 g/mol
mol AgNO3 : 30.1 g / 169.37 g/mo = 0.18 mol AgNO3
0.18 mol AgNO3 x 1 mol AgCl/moll AgNO3 = 0.18 mol AgCl
0.18 mol AgCL x 143.32 g / mol AgCl = 25.40 g AgCl