Question

Heavy metal ions like lead(II) can be precipitated from laboratory wastewater by adding sodium sulfide, Na2S. Will all the lead be removed from 11.2 mL of 7.10×10-3 M Pb(NO3)2 upon addition of 12.4 mL of 0.0117 M Na2S? If all the lead is removed, how many moles of lead is this? If not, how many moles of Pb remain?

Answer 1

Answer: The moles of lead removed is $7.95\times 10^{-5}mol$

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

• For sodium sulfide:

Molarity of sodium sulfide solution = 0.0117 M

Volume of solution = 12.4 mL = 0.0124 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

$0.0117M=\frac{\text{Moles of }Na_2S}{0.0124L}\\\\\text{Moles of }Na_2S=(0.0117mol/L\times 0.0124L)=1.451\times 10^{-4}mol$

• For lead nitrate:

Molarity of lead nitrate solution = $7.10\times 10^{-3}$ M

Volume of solution = 11.2 mL = 0.0112 L

Putting values in equation 1, we get:

$7.10\times 10^{-3}M=\frac{\text{Moles of }Pb(NO_3)_2}{0.0112L}\\\\\text{Moles of }Pb(NO_3)_2=(7.10\times 10^{-3}mol/L\times 0.0112L)=7.95\times 10^{-5}mol$

The chemical equation for the reaction of sodium sulfide and lead nitrate follows:

$Pb(NO_3)_2+Na_2S\rightarrow PbS+2NaNO_3$

By Stoichiometry of the reaction:

1 mole of lead nitrate reacts with 1 mole of sodium sulfide.

So, $7.95\times 10^{-5}mol$ will react with = $\frac{1}{1}\times 7.95\times 10^{-5}=7.95\times 10^{-5}mol$ of sodium sulfide.

As, given amount of sodium sulfide is more than the required amount. So, it is considered as an excess reagent.

Thus, lead nitrate is considered as a limiting reagent because it limits the formation of product.

As, all the lead from lead nitrate is getting converted to lead sulfide. So, all the lead (II) is removed from the solution.

By Stoichiometry of the reaction:

1 mole of lead nitrate produces 1 mole of lead sulfide.

So, $7.95\times 10^{-5}mol$ will produce = $\frac{1}{1}\times 7.95\times 10^{-5}=7.95\times 10^{-5}mol$ of lead sulfide.

Hence, the moles of lead removed is $7.95\times 10^{-5}mol$

Answer 2

Answer:

The answer to your question is: 0.0000784 moles of lead (II) there were in the solution.

Explanation:

Data

Pb(NO3)2       11.2 ml  = 0.0112 l      7.10 x 10 ⁻³ M

Na₂S               12.4 ml = 0.0124 l      0.0117 M

# of moles = ?

Process

Molarity = # moles / volume

# moles = Molarity x volume

For Pb(NO₃)₂

# moles = (7 x 10⁻³ ) / 0.0112)

             = 0.0000784

For Na₂S

# moles = (0.0117)(0.0124)

             = 0.00015

Balanced Reaction

                     Pb(NO₃)₂    +     Na₂S     ⇒      PbS     +    2NaNO₃

The proportion    Pb(NO₃)₂  to  Na₂S  is:  1 : 1

Then

                  1 mol of  Pb(NO₃)₂  -----------------  1 mol of Na₂S  

               0.0000784 mol        ----------------      x

                  x = ( 0.0000784 x 1) / 1

                 x = 0.0000784 mol of Na₂S

Then it was poured 0.00015 moles of Na₂S and reacted 0.0000784 moles

That means that all the Pb(NO₃)₂ reacted and there was an excess of Na₂S.