5 mL of 0.1 M Pb(NO_{3})_{2} is mixed with | vedclass

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(B) The balanced chemical equation for the reaction is:$Pb(NO_{3})_{2}(aq) + 2KI(aq) \rightarrow PbI_{2}(s) + 2KNO_{3}(aq)$Calculate the initial milli

Vedclass · Accepted Answer

$10^{-4} \ mol$

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(B) The balanced chemical equation for the reaction is:$Pb(NO_{3})_{2}(aq) + 2KI(aq) ightarrow PbI_{2}(s) + 2KNO_{3}(aq)$Calculate the initial millimoles of each reactant:Millimoles of $Pb(NO_{3})_{2} = 5 \ mL 	imes 0.1 \ M = 0.5 \ mmol$Millimoles of $KI = 10 \ mL 	imes 0.02 \ M = 0.2 \ mmol$According to the stoichiometry, $1 \ mol$ of $Pb(NO_{3})_{2}$ reacts with $2 \ mol$ of $KI$. Therefore, $0.2 \ mmol$ of $KI$ will react with $0.1 \ mmol$ of $Pb(NO_{3})_{2}$.Since $KI$ is the limiting reagent, the amount of $PbI_{2}$ formed depends on the amount of $KI$.From the stoichiometry, $2 \ mmol$ of $KI$ produces $1 \ mmol$ of $PbI_{2}$.So, $0.2 \ mmol$ of $KI$ will produce $0.1 \ mmol$ of $PbI_{2}$.Convert millimoles to moles:$0.1 \ mmol = 0.1 	imes 10^{-3} \ mol = 10^{-4} \ mol$.

$5 \ mL$ of $0.1 \ M \ Pb(NO_{3})_{2}$ is mixed with $10 \ mL$ of $0.02 \ M \ KI$. The amount of $PbI_{2}$ precipitated will be about

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