Question

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: What mass of precipitate will form if 1.50 L of highly concentrated Pb ( ClO 3 ) 2 is mixed with 0.200 L 0.300 M NaI

Answer 1

Answer: The mass of precipitate (lead (II) iodide) that will form is 13.83 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.300 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.300M=\frac{\text{Moles of NaI}}{0.200L}\\\\\text{Moles of NaI}=(0.300mol/L\times 0.200L)=0.06mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.06 moles of NaI will produce = $\frac{1}{2}\times 0.06=0.03mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.03 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.03mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.03mol\times 461.1g/mol)=13.83g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 13.83 grams