Upon mixing 50.0 mL of 0.1 M lead nitrate s | vedclass

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(A) The balanced chemical equation is: $3Pb(NO_3)_2 + Cr_2(SO_4)_3 	o 3PbSO_4 \downarrow + 2Cr(NO_3)_3$Initial millimoles $(mmol)$:$Pb(NO_3)_2 = 50 \

Vedclass · Accepted Answer

$0.005, 0.0084$

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(A) The balanced chemical equation is: $3Pb(NO_3)_2 + Cr_2(SO_4)_3 	o 3PbSO_4 \downarrow + 2Cr(NO_3)_3$Initial millimoles $(mmol)$:$Pb(NO_3)_2 = 50 \ mL 	imes 0.1 \ M = 5 \ mmol$$Cr_2(SO_4)_3 = 50 \ mL 	imes 0.05 \ M = 2.5 \ mmol$Determine the limiting reagent:According to the stoichiometry,$3 \ mmol$ of $Pb(NO_3)_2$ reacts with $1 \ mmol$ of $Cr_2(SO_4)_3$.For $5 \ mmol$ of $Pb(NO_3)_2$,the required $Cr_2(SO_4)_3 = \frac{1}{3} 	imes 5 = 1.667 \ mmol$.Since we have $2.5 \ mmol$ of $Cr_2(SO_4)_3$,$Pb(NO_3)_2$ is the limiting reagent.$(I)$ Moles of $PbSO_4$ formed:$3 \ mmol$ of $Pb(NO_3)_2$ produces $3 \ mmol$ of $PbSO_4$.$5 \ mmol$ of $Pb(NO_3)_2$ produces $5 \ mmol = 0.005 \ mol$ of $PbSO_4$.$(II)$ Concentration of $Cr_2(SO_4)_3$ left:Initial $mmol = 2.5 \ mmol$Reacted $mmol = 1.667 \ mmol$Left $mmol = 2.5 - 1.667 = 0.833 \approx 0.84 \ mmol$Total volume $= 50 \ mL + 50 \ mL = 100 \ mL$Concentration $= \frac{0.84 \ mmol}{100 \ mL} = 0.0084 \ M$

Upon mixing $50.0 \ mL$ of $0.1 \ M$ lead nitrate solution with $50 \ mL$ of $0.05 \ M$ chromic sulphate solution,precipitation of lead sulphate takes place. How many moles of lead sulphate are formed and what is the molar concentration of chromic sulphate left in the solution?

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