Question

B. AgNO3 + NaCl → AgCl + Nanoz.
How many grams of silver nitrate is required to precipitate 287 gm of silver
chloride?
(N=23, 0=16, CI = 35.5, Ag = 108)
​

Answer 1

340g of silver nitrate is required to precipitate 287 gm of silver

chloride

explanation : AgNO3 + NaCl ⇒ AgCl(↓) + NaNO3

here it is clear that, one mole of silver nitrate is required to precipitate one mole of silver chloride.

• molecular weight of AgNO3 = 170 g/mol
• molecular weight of AgCl = 143.5 g/mol

so, 170g of AgNO3 is required to precipitate 143.5 g of AgCl.

or, 170/143.5 × 287 g = 340g of AgNO3 is required to precipitate 287g of AgCl.

also read similar questions :

21.6g of silver coins is dissolved in hno3.when nacl is added to this solution all silver is precipitated as agcl The wi...

https://brainly.in/question/3860742

white precipitate of silver chloride is formed on adding a solution of sodium chloride to Silver Nitrate solution balanc...

https://brainly.in/question/2687499