How many grams of Pb(IO3)2(s) are precipitated when 27.0 mL of 3.00 M Pb(NO3)2(aq) are mixed with 25.0 mL of a 5.00 M KIO3(aq) solution?
To determine the number of grams of Pb(IO3)2(s) precipitated when the two solutions are mixed, we will use the concept of stoichiometry and the balanced equation for the reaction between Pb(NO3)2 and KIO3.
The balanced equation for the reaction is:
Pb(NO3)2 (aq) + 2 KIO3 (aq) -> Pb(IO3)2 (s) + 2 KNO3 (aq)
First, let's find the limiting reagent - the reactant that will be entirely consumed in the reaction. This will determine the amount of Pb(IO3)2(s) precipitated.
1. Calculate the number of moles of Pb(NO3)2:
mol(Pb(NO3)2) = Molarity(Pb(NO3)2) x Volume(Pb(NO3)2)
= 3.00 M x 0.0270 L
= 0.0810 moles
2. Calculate the number of moles of KIO3:
mol(KIO3) = Molarity(KIO3) x Volume(KIO3)
= 5.00 M x 0.0250 L
= 0.125 moles
3. Determine the stoichiometric ratio between Pb(NO3)2 and Pb(IO3)2 from the balanced equation. It is 1:1, meaning one mole of Pb(NO3)2 reacts to form one mole of Pb(IO3)2.
Since the stoichiometric ratio is 1:1, the limiting reagent is the reactant that has fewer moles. In this case, Pb(NO3)2 has 0.0810 moles, while KIO3 has 0.125 moles.
4. Calculate the moles of Pb(IO3)2 formed:
Since one mole of Pb(NO3)2 reacts to form one mole of Pb(IO3)2, the number of moles of Pb(IO3)2 formed will be equal to the number of moles of Pb(NO3)2 used.
moles of Pb(IO3)2 = mol(Pb(NO3)2)
= 0.0810 moles
Finally, we can calculate the mass of Pb(IO3)2 using its molar mass:
5. Calculate the mass of Pb(IO3)2:
Mass(Pb(IO3)2) = moles of Pb(IO3)2 x Molar mass(Pb(IO3)2)
The molar mass of Pb(IO3)2 can be calculated by adding the atomic masses of Pb, I, and O multiplied by their respective subscripts in the formula.
Molar mass(Pb(IO3)2) = (Atomic mass(Pb) + 2 x Atomic mass(I) + 6 x Atomic mass(O))
= (207.2 g/mol + 2 x 126.9 g/mol + 6 x 16.0 g/mol)
= 551.0 g/mol
Mass(Pb(IO3)2) = 0.0810 moles x 551.0 g/mol
= 44.5 grams
Therefore, when 27.0 mL of 3.00 M Pb(NO3)2(aq) is mixed with 25.0 mL of a 5.00 M KIO3(aq) solution, approximately 44.5 grams of Pb(IO3)2(s) will be precipitated.