Respuesta :
Answer: The moles of lead removed is [tex]7.95\times 10^{-5}mol[/tex]
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
- 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:
[tex]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[/tex]
- For lead nitrate:
Molarity of lead nitrate solution = [tex]7.10\times 10^{-3}[/tex] M
Volume of solution = 11.2 mL = 0.0112 L
Putting values in equation 1, we get:
[tex]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[/tex]
The chemical equation for the reaction of sodium sulfide and lead nitrate follows:
[tex]Pb(NO_3)_2+Na_2S\rightarrow PbS+2NaNO_3[/tex]
By Stoichiometry of the reaction:
1 mole of lead nitrate reacts with 1 mole of sodium sulfide.
So, [tex]7.95\times 10^{-5}mol[/tex] will react with = [tex]\frac{1}{1}\times 7.95\times 10^{-5}=7.95\times 10^{-5}mol[/tex] 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, [tex]7.95\times 10^{-5}mol[/tex] will produce = [tex]\frac{1}{1}\times 7.95\times 10^{-5}=7.95\times 10^{-5}mol[/tex] of lead sulfide.
Hence, the moles of lead removed is [tex]7.95\times 10^{-5}mol[/tex]
All the lead ions will not be removed and 6.502 ×10-8 moles of lead ions remain in solution.
The equation of the reaction is;
Pb(NO3)2(aq) + Na2S(aq) ---------> PbS(s) + 2NaNO3(aq)
Number of moles of Pb(NO3)2 = concentration × volume
Number of moles of Pb(NO3)2 = 11.2 mL/100 × 7.10×10-3 M = 7.952 ×10-8 moles
Number of moles of Na2S = 12.4 mL/1000 × 0.0117 M = 1.45 × 10-8 moles
Since the reaction is 1:1, 7.952 ×10-8 moles of Pb(NO3)2 should react with 7.952 ×10-8 moles of Na2S in order to remove all the Pb^2+ ions but we have only 1.45 × 10-8 moles of Na2S hence all the lead ions will not be removed.
Amount of lead ions remaining = 7.952 ×10-8 - 1.45 × 10-8 moles
= 6.502 ×10-8 moles of lead ions
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