What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.150 L of 0.190 M NaI? Assume the reaction goes to completion.
some? maybe a lot? perhaps nothing? I don't know what concentrated Pb(ClO3)2 means. The problem can't be worked unless the concentration is known.
To determine the mass of precipitate that will form, we need to first balance the chemical equation for the reaction between Pb(ClO3)2 and NaI.
The balanced chemical equation is:
Pb(ClO3)2 + 2 NaI → PbI2 + 2 NaClO3
From the balanced equation, we can see that 1 mole of Pb(ClO3)2 reacts with 2 moles of NaI to produce 1 mole of PbI2.
Next, we need to calculate the number of moles of Pb(ClO3)2 and NaI.
To find the number of moles of Pb(ClO3)2, we'll use its concentration and volume.
Concentrated Pb(ClO3)2 is not given in the problem, so we'll assume a molarity of 6.00 M (for example).
moles of Pb(ClO3)2 = concentration × volume
= 6.00 M × 1.50 L
= 9.00 moles
Next, we'll calculate the number of moles of NaI using its concentration and volume.
moles of NaI = concentration × volume
= 0.190 M × 0.150 L
= 0.0285 moles
Now that we have the moles of Pb(ClO3)2 and NaI, we can determine the limiting reagent. The limiting reagent is the reactant that is completely used up in the reaction and determines the maximum amount of product that can be formed.
From the balanced equation, we know that 1 mole of Pb(ClO3)2 reacts with 2 moles of NaI. This ratio indicates that 1 mole of Pb(ClO3)2 requires 2 moles of NaI for complete reaction.
Using the stoichiometry, if 1 mole of Pb(ClO3)2 reacts with 2 moles of NaI, then 9.00 moles of Pb(ClO3)2 will react with (2/1) × 9.00 = 18.00 moles of NaI.
Since we have only 0.0285 moles of NaI, which is less than the required 18.00 moles, NaI is the limiting reagent.
Now, let's calculate the moles of PbI2 formed based on the limiting reagent.
Since the stoichiometry of the balanced equation tells us that 1 mole of Pb(ClO3)2 produces 1 mole of PbI2, we can conclude that 9.00 moles of Pb(ClO3)2 will produce 9.00 moles of PbI2.
Therefore, the mass of PbI2 can be calculated using its molecular weight.
molecular weight of PbI2 = 207.2 g/mol
mass of PbI2 = moles of PbI2 × molecular weight of PbI2
= 9.00 moles × 207.2 g/mol
= 1864.8 g
Therefore, if the reaction goes to completion, a mass of 1864.8 grams of PbI2 will form.