Question

and the number of moles of pbcl2 produced from the complete reaction of pb(

mol pbcl2 = mol pb(no3)2×\frac{1 mol pbcl2}{1 mol pb(no3)2}

= 1.26×10^{-2} \underline{mol pb(no3)2}×\frac{1 mol}{1 mol pb}

= 1.26×10^{-2} mol pbcl2

therefore, the limiting reactant is pb(no3)2.

part b

determine the theoretical yield of pbcl2.
express your answer in grams to three significant figures.

g

Explanation:

Step1: Recall molar - mass formula

The molar mass of $PbCl_2$ is calculated as follows: The molar mass of $Pb$ is $207.2\ g/mol$ and the molar mass of $Cl$ is $35.45\ g/mol$. Since there are 2 $Cl$ atoms in $PbCl_2$, the molar mass of $Cl_2$ is $2\times35.45 = 70.9\ g/mol$. The molar mass of $PbCl_2$, $M = 207.2+70.9=278.1\ g/mol$.

Step2: Use the mole - mass relationship

We know from part A that the number of moles of $PbCl_2$ produced (based on the limiting reactant $Pb(NO_3)_2$) is $n = 1.26\times 10^{-2}\ mol$. According to the formula $m=n\times M$, where $m$ is the mass, $n$ is the number of moles, and $M$ is the molar mass. Substitute $n = 1.26\times 10^{-2}\ mol$ and $M = 278.1\ g/mol$ into the formula: $m=(1.26\times 10^{-2}\ mol)\times278.1\ g/mol$.

Step3: Calculate the mass

$m=(1.26\times 10^{-2})\times278.1 = 3.50406\ g$. Rounding to three significant figures, $m = 3.50\ g$.

Answer:

$3.50\ g$