What weight of AgCl will be precipitated when a solution containing 4 . 77 g of NaCl is added to a solution of 5 . 77 g of `AgNO_(3)` ?
`(Na = 23, Cl = 35 . 5, Ag = 108 , N = 14 and O = 16)`

Explanation

`NaCl + Ag NO_(3) to Na NO_(3) + Ag Cl`
No . Of moles of `NaCl = ( 4 . 77)/( 58 . 5) = 0 . 08154`
No of moles of `ag NO_(3) = ( 5 . 77) /( 170) = 0 . 0 3394`
Since no. of moles of `AgNO_(3)` is less then that of NaCl the whole of `AgNO_(3)` shall convert into AgCl `AgNO_(3)` is a limiting reagent). Applying thus the POAC for Ag atoms as the Ag atoms are conserved .
moles of Ag in `AgNO_(3) ` = moles of Ag in AgCl
` 1 xx` moles of `AgNO_(3)= 1 xx ` moles of AgCl
`(wt. of AgNO_(3))/( mol. wt. of AgNO_(3)) = (wt.of AgCl)/( mol.wt.of AgCl)`
Wt. of AgCl = ` 0 . 0 3394 xx 143.5 = 4.87` g