equation for the reaction of aqueous Pb(ClO3)₂ with aqueous Nal. Include phases. chemical equation: What mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO3)₂ is mixed with 0.500 L 0.300 M Nal? Assume the reaction goes to completion. mass of precipitate: bo

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Chapter1: Chemical Foundations
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**Chemical Reaction and Precipitate Formation**

**Prompt:**

1. *Write the balanced equation for the reaction of aqueous Pb(ClO₃)₂ with aqueous NaI. Include phases.*

   **Chemical equation:** 
   \[ \text{Pb(ClO}_3\text{)}_2(aq) + 2\text{NaI}(aq) \rightarrow \text{PbI}_2(s) + 2\text{NaClO}_3(aq) \]

2. *What mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO₃)₂ is mixed with 0.500 L of 0.300 M NaI? Assume the reaction goes to completion.*

   **Mass of precipitate:** 
   \[ \underline{\hspace{4cm}} \, \text{g} \]

This exercise involves calculating the mass of lead(II) iodide (\(\text{PbI}_2\)) precipitate that forms when lead(II) chlorate (\(\text{Pb(ClO}_3\text{)}_2\)) reacts with sodium iodide (\(\text{NaI}\)). To solve this, use stoichiometry based on the balanced chemical equation.
Transcribed Image Text:**Chemical Reaction and Precipitate Formation** **Prompt:** 1. *Write the balanced equation for the reaction of aqueous Pb(ClO₃)₂ with aqueous NaI. Include phases.* **Chemical equation:** \[ \text{Pb(ClO}_3\text{)}_2(aq) + 2\text{NaI}(aq) \rightarrow \text{PbI}_2(s) + 2\text{NaClO}_3(aq) \] 2. *What mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO₃)₂ is mixed with 0.500 L of 0.300 M NaI? Assume the reaction goes to completion.* **Mass of precipitate:** \[ \underline{\hspace{4cm}} \, \text{g} \] This exercise involves calculating the mass of lead(II) iodide (\(\text{PbI}_2\)) precipitate that forms when lead(II) chlorate (\(\text{Pb(ClO}_3\text{)}_2\)) reacts with sodium iodide (\(\text{NaI}\)). To solve this, use stoichiometry based on the balanced chemical equation.
Expert Solution
Step 1

mass = mole × molar mass 

In this reaction, 1 mole Pb(ClO3)2 react with 2 mole  NaI to form 1 mole PbI2 (s) precipitate.

mole of precipitate = mole of NaI2

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