how much silver chloride will be formed by mixing 120 g of silver nitrate with a solution of 52.0 g of NaCL?
NaCl + AgNO3 -> AgCl + NaNO3
Answers
Answer:
sorry
Explanation:
I really didn't know the correct explaination and answer
Answer:
12.8
Explanation:
Q: How much AgCl will be formed by mixing 120g of AgNO3 with a solution of 52g NaCl
AgNO3 + NaCl → AgCl + NaNO3
Given Data:
Mass of AgNO3= 120g
Masss of NaCl= 52g
Required:
Mass of AgCl=?
Solution:
Mass of AgNO3 (given)= 120g
Molar mass of AgNO3 = 169g
No of moles of AgNO3=120/169
No of moles of AgNO3= 0.71moles
Mass of NaCl (given) = 52g
Molar mass of NaCl = 58
No of moles of NaCl =52/58
No of moles of NaCl = 0.9 moles
Stochiometrically:
1 mole of AgNO3 = 1 mole of AgCl
0.71 mole of AgNO3 = 0.71 moles of AgCl
1 mole of Nacl = 1 mole of AgCl
0.9 moles of NaCl = 0.9 moles of AgCl
As NaCl is producing lesser moles of product , so it is the limiting reactant and produces 0.9 moles of AgCl
Mass of AgCl = Molees x Atomic Mass
Mass of AgCl = 0.9 x 143
Mass of AgCl = 128.7