A sample of pure lead weighing `2.07 g` is dissolved in nitric acid to give a solution of lead nitrate. This solution is treated with hydrochloric acid, chlorine gas and ammonium chloride. The result is a precipitate of `(NH_(4))_(2) PbCl_(6)`. What is the maximum weight of this product that could be obtained form the lead sample?

Explanation

Eq. of `Pb=Eq. "of" Pb(NO_(3))_(2)`
`=Eq."of"(NH_(4))_(2)PbCl_(6)`
`(2.07)/(207//2)=(w)/(456//2)`
`:. Wt. "of" (NH_(4))_(2)PbCl_(6), w= 4.56 g`