What weight of `AgCl` will be precipitated when a solution containing `4.77 g NaCl` is added to a solution of `5.77 g` of `AgNO_(3)`.

Answer

C, D

Explanation

`underset({:("Initial"),(Left):})( ) underset({:((5.77)/(170) = 0.033),(0):})(AgNO_(3)) + underset({:((4.77)/(58.5) = 0.08),(0.08 - 0.033),(= 0.048):})(NaCl)rarrunderset({:(0),(0.033):})(AgCl) + underset({:(0),(0.033):})(NaNO_(3))`
`[Mw of AgNO_(3) = 108 + 14 + 16 xx 3 = 170g]`
`[Mw of NaCl = 23 + 35.5 = 48.5 g]`
0.033 moles of `AgCl` is formed.
Weight of `AgCL = 0.033 xx 143.5 = 4.7355 g`
Weight `= "Moles" xx Mw, Mw of AgCl = 108 + 35.5 = 143.5 g`