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 0.200 M NaI ? Assume the reaction goes to completion.

Answer 1

Answer: The mass of lead iodide produced is 9.22 grams

Step-by-step explanation:

To calculate the molarity of solution, we use the equation:

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

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

`0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L* 0.200L)=0.04moles`

The chemical equation for the reaction of NaI and lead chlorate follows:

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

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with =
`(1)/(2)* 0.04=0.02mol` of lead iodide

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

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

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

`0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol* 461g/mol)=9.22g`

Hence, the mass of lead iodide produced is 9.22 grams