What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.750 L of 0.290 M NaI? Assume the reaction goes to completion.
This is a limiting reagent problem; I know that because amounts for BOTH reactants are given. Here is a worked example of a limiting reagent problem. It will work all of them for you so print these instructions.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
19.4g
To determine the mass of precipitate formed, we first need to identify the balanced chemical equation for the reaction between lead(II) perchlorate (Pb(ClO3)2) and sodium iodide (NaI).
The balanced chemical equation for the reaction is:
Pb(ClO3)2 + 2NaI → PbI2 + 2NaClO3
From the balanced equation, we can see that 1 mole of Pb(ClO3)2 reacts with 2 moles of NaI to form 1 mole of PbI2.
Next, let's calculate the number of moles of Pb(ClO3)2 and NaI we have.
Given:
Volume of Pb(ClO3)2 solution = 1.50 L
Concentration of NaI solution = 0.290 M
Volume of NaI solution = 0.750 L
Using the formula:
moles = concentration x volume
Moles of Pb(ClO3)2 = concentration of Pb(ClO3)2 x volume of Pb(ClO3)2
= unknown concentration (let's call it x) x 1.50 L
= 1.50x
Moles of NaI = concentration of NaI x volume of NaI
= 0.290 M x 0.750 L
= 0.2175 moles
According to the balanced equation, the ratio between Pb(ClO3)2 and PbI2 is 1:1. This means that the number of moles of PbI2 formed will also be 1.50x.
Now, we need to look at the stoichiometry of the reaction to find out how many moles of Pb(ClO3)2 are required to form 1 mole of PbI2. In this case, it is a 1:1 ratio.
Since our reaction goes to completion, we know that all the NaI will react with Pb(ClO3)2.
Using stoichiometry, we can see that:
1.50x moles of Pb(ClO3)2 will produce 1.50x moles of PbI2
Therefore, the number of moles of PbI2 formed is also 1.50x.
Finally, we can calculate the mass of PbI2 using the molar mass of PbI2, which is 461.01 g/mol.
Mass of PbI2 = moles of PbI2 x molar mass of PbI2
= 1.50x moles x 461.01 g/mol
= 691.52x g
To determine the mass of precipitate formed, we need to know the concentration of Pb(ClO3)2, which is missing from the given information. Without this concentration, we cannot determine the exact mass of precipitate that will form.