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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 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 = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

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

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

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

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

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ 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 = 0.26 moles

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

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago

What is the uses of plastic?

Alex Ar [27]

Answer:

Plastic is used across almost every sector, including to produce packaging, in building and construction, in textiles, consumer products, transportation, electrical and electronics and industrial machinery.

7 0

2 years ago

Read 2 more answers

Which of the following additions to alkenes occur(s) specifically in an syn fashion?

BlackZzzverrR [31]

Answer:

E) A, B, and C

Explanation:

Syn addition refers to the addition of two substituents on the same face or side of a double bond. This differed from anti addition which a occurs across opposite face of the double bond.

Hydrogenation, hydroboration and dihydroxylation all involve syn addition to the double bond, hence the answer chosen above.

8 0

3 years ago

A sample of Neon is in a sealed container held under isothermic conditions. The initial pressure and volume are 2.7 atm and 4.5

Marina CMI [18]

Answer:

The final volume in mL is 7.14 mL or 7.1 mL.

Explanation:

1.Use Boyle's Law( $P_{1} V_{1}= P_{2} V_{2}$ ). Re-arrange to solve for $V_{2}$ for the final volume.