1.

A solution of lead nitrate prepared by dissolving 2.07 g ofpure lead innitricacid was treated with HCI, Cl_(2) gas and NH_(4) Cl . Whatwill be the maximumweightof (NH_(4))_(2) Pb Cl_(6) so produced ?

Answer»

Solution :We have, `underset(2.07 g ) (Pb) to Pb(NO_(3))_(2) underset (Cl_(2)NH_(4)Cl)overset(HCI)to (NH_(4))_(2) Pb Cl_(6) `
Now , for MAXIMUM yield of `(NH_(4))_(2) PbCl_(6)`
molesof Pb in the reactant = moles of Pb in `(NH_(4))_(2) Pb Cl_(6)`
` 1 xx ` moles of `(NH_(4))_(2) Pb Cl_(6)`
or `(2 . 07)/( 207) = ("maximum wt. of"(NH_(4))_(2) Pb Cl_(6))/(456)""[(NH_(4))_(2) Pb Cl_(6) = 456]`
`:.` maximumwt. of `(NH_(4))_(2)PbCl_(6) = 4 . 56 g`


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