Answer to Question #280745 in General Chemistry for sarah

Question

You add 3.00 g of Pb(NO₃)₂ (M.W. = 331.2 g/mole) to 75.0 mL of 0.100 M HCl, leading to formation of 0.873 g PbCl₂ (M.W. = 278.1 g/mole) as a precipitate.

a) Which molecule is the limiting reactant?

 

b) What is the percent yield of PbCl₂? 

Answer 1

The balanced equation is below:

Pb(NO₃)₂ + 2HCl --> PbCl₂ + 2HNO₃

a)

$n(Pb(NO_3)_2)=3.00\ g\times\frac{1\ mol}{331.2\ g}=0.009058\ mol$

$n(HCl)=0.0750\ L\times\frac{0.100\ mol}{1\ L}=0.00750\ mol$

According to the equation, 1 mole of Pb(NO₃)₂ reacts with 2 moles of HCl. Actually, there are fewer moles HCl than Pb(NO₃)₂, hence HCl is the limiting reactant.

b)

$Theoretical\ yield\ PbCl_2=0.00750\ mol(HCl)\times\frac{1\ mol(PbCl_2)}{2\ mol(HCl)}\times\frac{278.1\ g(PbCl_2)}{1\ mol(PbCl_2)}=1.043\ g$

$\%\ yield=\frac{actual\ yield}{theoretical\ yield}\times100\%=83.7\%$

Answer: 83.7%