If a solution containing 108.9 g of mercury(II) perchlorate is allowed to react completely with a solution containing 16.642 g of sodium sulfide, how many grams of solid precipitate will be formed? mass: How many grams of the reactant in excess will remain after the reaction? mass: Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (0) for the number of moles. mol Hg2+: Question Source: McQuarrie, Rock, And Gallogly 4e- General Chemsitry Publisher: University Science ooks

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**Topic: Stoichiometry and Precipitation Reactions**

If a solution containing 108.9 g of mercury(II) perchlorate is allowed to react completely with a solution containing 16.642 g of sodium sulfide, how many grams of solid precipitate will be formed?

- **Mass of precipitate:**
  - (Input Box)
  - **g**

How many grams of the reactant in excess will remain after the reaction?

- **Mass of excess reactant:**
  - (Input Box)
  - **g**

Assuming complete precipitation, how many moles of each ion remain in the solution? If an ion is no longer in solution, enter a zero (0) for the number of moles.

- **Hg²⁺:**
  - (Input Box)
  - **mol**

_Question Source: McQuarrie, Rock, and Gallogly 4e - General Chemistry_ | _Publisher: University Science Books_

---

**Explanation of the problem:**
- This problem involves determining the mass of a solid precipitate formed from a chemical reaction between mercury(II) perchlorate and sodium sulfide.
- The first step is to determine how many grams of the solid precipitate are produced.
- Next, it is necessary to determine how much of the reactant is left unreacted.
- Finally, the moles of specific ions remaining in the solution are calculated to complete the analysis.

Use appropriate stoichiometric calculations based on the balanced chemical equation to solve each part of the problem. When entering values, ensure they are derived from precise and accurate computations based on the given masses.

_Copyright © 2011-2021 Sapling Learning, Inc._
Transcribed Image Text:**Topic: Stoichiometry and Precipitation Reactions** If a solution containing 108.9 g of mercury(II) perchlorate is allowed to react completely with a solution containing 16.642 g of sodium sulfide, how many grams of solid precipitate will be formed? - **Mass of precipitate:** - (Input Box) - **g** How many grams of the reactant in excess will remain after the reaction? - **Mass of excess reactant:** - (Input Box) - **g** Assuming complete precipitation, how many moles of each ion remain in the solution? If an ion is no longer in solution, enter a zero (0) for the number of moles. - **Hg²⁺:** - (Input Box) - **mol** _Question Source: McQuarrie, Rock, and Gallogly 4e - General Chemistry_ | _Publisher: University Science Books_ --- **Explanation of the problem:** - This problem involves determining the mass of a solid precipitate formed from a chemical reaction between mercury(II) perchlorate and sodium sulfide. - The first step is to determine how many grams of the solid precipitate are produced. - Next, it is necessary to determine how much of the reactant is left unreacted. - Finally, the moles of specific ions remaining in the solution are calculated to complete the analysis. Use appropriate stoichiometric calculations based on the balanced chemical equation to solve each part of the problem. When entering values, ensure they are derived from precise and accurate computations based on the given masses. _Copyright © 2011-2021 Sapling Learning, Inc._
### Determining Remaining Ions Post-Precipitation

Assuming complete precipitation, calculate the number of moles of each ion that remain in solution. If an ion is no longer in solution, enter zero (0) for the number of moles.

**Hg\(^2+\):**
\[ \_\_\_\_\_\_\_\_\_ \text{ mol} \]

**ClO\(_4^-\):**
\[ \_\_\_\_\_\_\_\_\_ \text{ mol} \]

**Na\(^+\):**
\[ \_\_\_\_\_\_\_\_\_ \text{ mol} \]

**S\(^2-\):**
\[ \_\_\_\_\_\_\_\_\_ \text{ mol} \]

This exercise requires assessing the scenario where complete precipitation occurs, meaning certain ions will form a solid precipitate and will no longer be present in the aqueous solution. The result should be entered for each ion accordingly.

---

**Source:**
Question Source: McQuarrie, Rock, and Gallogly 4e - General Chemistry | Publisher: University Science Books

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Transcribed Image Text:### Determining Remaining Ions Post-Precipitation Assuming complete precipitation, calculate the number of moles of each ion that remain in solution. If an ion is no longer in solution, enter zero (0) for the number of moles. **Hg\(^2+\):** \[ \_\_\_\_\_\_\_\_\_ \text{ mol} \] **ClO\(_4^-\):** \[ \_\_\_\_\_\_\_\_\_ \text{ mol} \] **Na\(^+\):** \[ \_\_\_\_\_\_\_\_\_ \text{ mol} \] **S\(^2-\):** \[ \_\_\_\_\_\_\_\_\_ \text{ mol} \] This exercise requires assessing the scenario where complete precipitation occurs, meaning certain ions will form a solid precipitate and will no longer be present in the aqueous solution. The result should be entered for each ion accordingly. --- **Source:** Question Source: McQuarrie, Rock, and Gallogly 4e - General Chemistry | Publisher: University Science Books *Sapling Learning, Inc. © 2011-2021* For help or additional queries, please refer to the following links: - [Contact Us](#) - [Careers](#) - [Privacy Policy](#) - [Terms of Use](#) - [Help](#)
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