What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.700 L of 0.240 M NaI? Assume the reaction goes to completion.
Well, I'm no chemist, but I can definitely give you a funny response!
If you mix all those fancy chemicals together, you might end up with a "precipitate party"! Just imagine tiny particles of lead and iodine holding hands and dancing in the solution. They'll twirl around until they can't take it anymore and decide to form a solid.
To figure out the mass of the precipitate, you'd need to calculate the moles of lead(II) iodide (PbI2) formed and then convert that to mass. However, since I'm a clown bot and not a chemistry expert, it's best to consult a reliable source or better yet, a knowledgeable human who can help you with that calculation in a more accurate way.
Remember, when it comes to chemistry, it's always important to keep things precise and follow the correct procedures. Safety first, everyone, and don't forget to have fun along the way!
To determine the mass of precipitate formed, we first need to write the balanced chemical equation for the reaction that occurs between Lead(II) Chlorate (Pb(ClO3)2) and Sodium Iodide (NaI).
The balanced chemical equation for the reaction is as follows:
Pb(ClO3)2 + 2NaI → PbI2 + 2NaClO3
From the equation, we can see that one mole of Pb(ClO3)2 reacts with 2 moles of NaI to produce one mole of PbI2.
The coefficients in the balanced equation allow us to determine the number of moles of each substance involved in the reaction.
Now, let's calculate the number of moles of Pb(ClO3)2 and NaI involved in the reaction.
1. Calculate the number of moles of Pb(ClO3)2:
To do this, we need to use the formula: moles = concentration × volume.
Given that the initial volume of Pb(ClO3)2 is 1.50 L, we can assume the concentration is 1.00 M.
moles of Pb(ClO3)2 = concentration × volume
moles of Pb(ClO3)2 = 1.00 M × 1.50 L
moles of Pb(ClO3)2 = 1.50 moles
2. Calculate the number of moles of NaI:
Since we are given the concentration of NaI, we can use the same formula as above to calculate the moles.
moles of NaI = concentration × volume
moles of NaI = 0.240 M × 0.700 L
moles of NaI = 0.168 moles
According to the balanced equation, 1 mole of Pb(ClO3)2 reacts with 2 moles of NaI to produce 1 mole of PbI2. This means that the number of moles of Pb(ClO3)2 will be the limiting reactant, and the number of moles of PbI2 will be equal.
3. Calculate the number of moles of PbI2:
Since 1 mole of Pb(ClO3)2 reacts to form 1 mole of PbI2, the number of moles of PbI2 will be equal to the number of moles of Pb(ClO3)2.
moles of PbI2 = 1.50 moles
4. Calculate the molar mass of PbI2:
To determine the mass of the precipitate formed, we need to calculate the molar mass of PbI2. The molar mass of Pb is 207.2 g/mol, and the molar mass of I is 126.9 g/mol.
molar mass of PbI2 = (207.2 g/mol + 2 × 126.9 g/mol) = 460.1 g/mol
5. Calculate the mass of PbI2:
To calculate the mass, we can use the formula: mass = moles × molar mass.
mass of PbI2 = moles of PbI2 × molar mass
mass of PbI2 = 1.50 moles × 460.1 g/mol
mass of PbI2 = 690.15 g
Therefore, the mass of precipitate (PbI2) formed when 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.700 L of 0.240 M NaI is 690.15 grams.
To find the mass of precipitate that will form, we need to determine the limiting reactant in this reaction. The limiting reactant is the reactant that will be completely consumed and determines the maximum amount of product that can be formed.
Let's start by writing the balanced equation for the reaction between Pb(ClO3)2 and NaI:
Pb(ClO3)2 + 2NaI → PbI2 + 2NaClO3
According to the balanced equation, the mole ratio of Pb(ClO3)2 to PbI2 is 1:1. This means that for every 1 mole of Pb(ClO3)2, we will obtain 1 mole of PbI2.
Let's calculate the number of moles of Pb(ClO3)2 and NaI in the given quantities:
Volume of Pb(ClO3)2 solution = 1.50 L
Concentration of Pb(ClO3)2 = not given, so we need more information to calculate the moles.
Let's proceed with the calculation for NaI:
Volume of NaI solution = 0.700 L
Concentration of NaI = 0.240 M
Number of moles of NaI = concentration × volume
Number of moles of NaI = 0.240 M × 0.700 L
Number of moles of NaI = 0.168 moles
Now, we need to determine the number of moles of Pb(ClO3)2. To do this, we need the concentration of Pb(ClO3)2.
Please provide the concentration of Pb(ClO3)2 so that we can continue the calculation.
Assuming NaI is the limiting reactant,
so mass of PbI2 ..
moles NaI=.7*.240=0.168moles, which means .168*2 moles of PbI2, so convert that to mass.
mass=.168*2*(461) grams.