How many grams of lead(II) iodate, Pb(IO3)2 (formula weight = 557.0 g/mol), are precipitated when 3.20 × 102 mL of 0.285 M Pb(NO3)2(aq) are mixed with 386 mL of 0.512 M NaIO3(aq) solution? The unbalanced equation of reaction is:
Pb(NO3)2(aq) + NaIO3(aq) → Pb(IO3)2(s) + NaNO3(aq)
50.8
Balance the equation.
Calculate moles of each from moles = M x L.
Determine the limiting reagent.
moles Pb(IO3)2 x molar mass = grams.
I note there is an excess of NaIO3 present which will reduce the solubility but I don't think you need to consider that.
From
To determine the number of grams of lead(II) iodate precipitated, we need to use stoichiometry and the given information.
1. Determine the moles of Pb(NO3)2:
- Given: Volume of Pb(NO3)2 = 3.20 × 102 mL = 0.320 L
- Concentration of Pb(NO3)2 = 0.285 M
- Moles of Pb(NO3)2 = Concentration × Volume
= 0.285 mol/L × 0.320 L
= 0.0912 mol
2. Determine the moles of NaIO3:
- Given: Volume of NaIO3 = 386 mL = 0.386 L
- Concentration of NaIO3 = 0.512 M
- Moles of NaIO3 = Concentration × Volume
= 0.512 mol/L × 0.386 L
= 0.197312 mol
3. Use stoichiometry to find the moles of Pb(IO3)2:
- From the balanced equation: 1 mole of Pb(NO3)2 reacts with 1 mole of Pb(IO3)2
- Moles of Pb(IO3)2 = Moles of Pb(NO3)2 (according to stoichiometry)
= 0.0912 mol
4. Convert moles of Pb(IO3)2 to grams:
- Given: Formula weight of Pb(IO3)2 = 557.0 g/mol
- Mass = Moles × Formula weight
= 0.0912 mol × 557.0 g/mol
= 50.7824 g
Therefore, approximately 50.7824 grams of lead(II) iodate, Pb(IO3)2, are precipitated in the reaction.