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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? |
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Answer» 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` |
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