How many moles of $P b {(N O_{3})}_{2}$ $Pb(NO_3)_2$ are required if 12 moles of $A l {(N O_{3})}_{3}$ $Al(NO_3)_3$ are produced?

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How many moles of $P b {(N O_{3})}_{2}$ $Pb(NO_3)_2$ are required if 12 moles of $A l {(N O_{3})}_{3}$ $Al(NO_3)_3$ are produced?

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Answer 1 · 2016-08-20T18:11:00

$3 P b {(N O_{3})}_{2} (a q) + 3 A l_{2} {(S O_{4})}_{3} (a q) \to 3 P b S O_{4} (s) ⏐ ↓ + 6 A l {(N O_{3})}_{3} (a q)$ $3Pb(NO_3)_2(aq) + 3Al_2(SO_4)_3(aq)rarr 3PbSO_4(s)darr + 6Al(NO_3)_3(aq)$

Explanation:

So if $12 \cdot m o l$ $12*mol$ of aluminum nitrate are produced, clearly, from the stoichiometry of the given reaction, $6$ $6$ $m o l$ $mol$ of $lead nitrate$ $"lead nitrate"$ were required. But don't trust my arithmetic!

I chose aluminum sulfate because it is a fact that both lead nitrate and aluminum sulfate are reasonably soluble in aqueous solution, whereas lead sulfate is fairly insoluble. Lead sulfate should thus precipitate out. Lead can thus be separated from aluminum with this metathesis reaction.

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How many moles of $P b {(N O_{3})}_{2}$ $Pb(NO_3)_2$ are required if 12 moles of $A l {(N O_{3})}_{3}$ $Al(NO_3)_3$ are produced?

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How many moles of Pb(NO3)2 Pb(NO_3)_2 are required if 12 moles of Al(NO3)3 Al(NO_3)_3 are produced?

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How many moles of $P b {(N O_{3})}_{2}$ $Pb(NO_3)_2$ are required if 12 moles of $A l {(N O_{3})}_{3}$ $Al(NO_3)_3$ are produced?