Question

missed this? watch kcv: reaction stoichiometry, we: limiting reactant and theoretical yield, we solution stoichiometry; read sections 4.3 and 5.3. you can click on the review link to access the section in your e - text. a 26.4 ml sample of a 1.74 m potassium chloride solution is mixed with 14.5 ml of a 0.870 m lead(ii) nitrate solution and this precipitation reaction occurs: 2kcl(aq)+pb(no3)2(aq)→pbcl2(s)+2kno3(aq). the solid pbcl2 is collected, dried, and found to have a mass of 2.47 g. determine the limiting reactant, the theoretical yield, and the percent yield. part a determine the limiting reactant. express your answer as a chemical formula. view available hint(s) part b complete previous part(s) part c complete previous part(s) provide feedback

Explanation:

Step1: Calculate moles of KCl

Use the formula $n = M\times V$ (where $n$ is moles, $M$ is molarity and $V$ is volume in L).
$V_{KCl}=26.4\ mL = 0.0264\ L$, $M_{KCl}=1.74\ M$
$n_{KCl}=1.74\ mol/L\times0.0264\ L = 0.045936\ mol$

Step2: Calculate moles of $Pb(NO_3)_2$

$V_{Pb(NO_3)_2}=14.5\ mL=0.0145\ L$, $M_{Pb(NO_3)_2}=0.870\ M$
$n_{Pb(NO_3)_2}=0.870\ mol/L\times0.0145\ L = 0.012615\ mol$

Step3: Determine mole - ratio from balanced equation

From $2KCl(aq)+Pb(NO_3)_2(aq)
ightarrow PbCl_2(s)+2KNO_3(aq)$, the mole - ratio of $KCl$ to $Pb(NO_3)_2$ is $2:1$.

Step4: Calculate moles of $Pb(NO_3)_2$ needed to react with $KCl$

If all of $KCl$ reacts, the moles of $Pb(NO_3)_2$ required is $\frac{n_{KCl}}{2}=\frac{0.045936\ mol}{2}=0.022968\ mol$. But we have only $0.012615\ mol$ of $Pb(NO_3)_2$.

Step5: Identify the limiting reactant

Since the amount of $Pb(NO_3)_2$ available is less than the amount required to react with all of $KCl$, $Pb(NO_3)_2$ is the limiting reactant.

Answer:

$Pb(NO_3)_2$