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 30.0 mL of a 5.00 M KIO3(aq) solution?
3.00M(.o27)(1mol Pb(IO3)2)/1mol Pb(NO3)2)(557/1)
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
Pb(NO3)2 + 2KIO3 ==> Pb(IO3)2 + 2KNO3
mols Pb(NO3)2 = M x L = ?
mols KIO3 = M x L = ?
Using the coefficients in the balanced equation, convert mols Pb(NO3)2 to mols Pb(IO3)2 and I think you have done that.
Do the same for mols KIO3 to mols Pb(IO3)2. It is likely the two answers will not be the same which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
Using the smaller value convert mols Pb(IO3)2 to g by grams = mols x molar mass. That is the theoretical yield for Pb(IO3)2 and it assumes everything goes at 100% (no losses).
45.1
To calculate the number of grams of Pb(IO3)2(s) precipitated, we need to use the stoichiometry of the balanced chemical equation to determine the number of moles of Pb(IO3)2.
The balanced chemical equation for the reaction between Pb(NO3)2 and KIO3 is:
Pb(NO3)2 + 2 KIO3 → Pb(IO3)2 + 2 KNO3
From the equation, we can see that one mole of Pb(NO3)2 reacts with one mole of Pb(IO3)2.
First, let's calculate the number of moles of Pb(NO3)2 present in the solution:
Number of moles of Pb(NO3)2 = concentration of Pb(NO3)2 × volume of Pb(NO3)2 solution
Number of moles of Pb(NO3)2 = 3.00 M × 0.0270 L (converted mL to L) = 0.081 mol
Next, let's calculate the number of moles of Pb(IO3)2 that will be formed. Since the stoichiometric ratio between Pb(NO3)2 and Pb(IO3)2 is 1:1, the number of moles of Pb(IO3)2 will be the same as the number of moles of Pb(NO3)2.
Number of moles of Pb(IO3)2 = 0.081 mol
To convert the number of moles of Pb(IO3)2 to grams, we need to know the molar mass of Pb(IO3)2. The molar mass of Pb(IO3)2 is calculated as follows:
Molar mass of Pb = 207.2 g/mol
Molar mass of I = 126.9 g/mol
Molar mass of O = 16.0 g/mol (there are three oxygen atoms in the formula)
Molar mass of Pb(IO3)2 = (207.2 g/mol) + (2 * 126.9 g/mol) + (6 * 16.0 g/mol) = 573.0 g/mol
Now, let's calculate the mass of Pb(IO3)2 that will be precipitated:
Mass of Pb(IO3)2 = Number of moles of Pb(IO3)2 × Molar mass of Pb(IO3)2
Mass of Pb(IO3)2 = 0.081 mol × 573.0 g/mol = 46.513 g
Therefore, approximately 46.5 grams of Pb(IO3)2 will be precipitated when 27.0 mL of 3.00 M Pb(NO3)2(aq) are mixed with 30.0 mL of a 5.00 M KIO3(aq) solution.