Question #2832
Calculate the mass of precipitate that forms when 250 mL of an aqueous solution containing 35.0g of lead (2) nitrate reacts with excess sodium iodide solution by the following reaction.
Pb(NO[sub]3[/sub])[sub]2[/sub] (aq) + 2NaI (aq) = 2NaNO[sub]3[/sub] (aq) + PbI[sub]2[/sub] (s)
Pb(NO[sub]3[/sub])[sub]2[/sub] (aq) + 2NaI (aq) = 2NaNO[sub]3[/sub] (aq) + PbI[sub]2[/sub] (s)
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#339914 on Dec 2023
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