-What volume of .500 M Pb(NO3)2 is required to react completely with 18.25 L of .625 M Na2SO4?
To find the volume of .500 M Pb(NO3)2 required to react completely with 18.25 L of .625 M Na2SO4, we can use the balanced chemical equation to determine the stoichiometry of the reaction.
The balanced chemical equation for the reaction between Pb(NO3)2 and Na2SO4 is:
Pb(NO3)2 + Na2SO4 -> PbSO4 + 2NaNO3
From the balanced equation, we can see that 1 mole of Pb(NO3)2 reacts with 1 mole of Na2SO4 to produce 1 mole of PbSO4.
First, let's calculate the number of moles of Na2SO4 in 18.25 L of .625 M Na2SO4.
Number of moles = Volume (in L) × Concentration (in M)
Number of moles of Na2SO4 = 18.25 L × 0.625 M = 11.40625 moles
Since the stoichiometry of the reaction is 1:1 between Pb(NO3)2 and Na2SO4, the number of moles of Pb(NO3)2 required to react completely with the given amount of Na2SO4 is also 11.40625 moles.
Now, let's find the volume of 0.500 M Pb(NO3)2 required to contain 11.40625 moles of Pb(NO3)2.
Volume (in L) = Number of moles / Concentration (in M)
Volume of 0.500 M Pb(NO3)2 = 11.40625 moles / 0.500 M = 22.8125 L
Therefore, the volume of 0.500 M Pb(NO3)2 required to react completely with 18.25 L of 0.625 M Na2SO4 is approximately 22.8125 L.