Aqueous hydrochloric acid (HCI) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H₂O). Suppose 32.4 g of hydrochloric acid is mixed with 26. g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits. 0 ? xto X S
Aqueous hydrochloric acid (HCI) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H₂O). Suppose 32.4 g of hydrochloric acid is mixed with 26. g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits. 0 ? xto X S
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Chemical Reaction Problem
#### Problem Statement
Aqueous hydrochloric acid (HCl) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H₂O).
Suppose 32.4 g of hydrochloric acid is mixed with 26.0 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits.
#### Instruction
Enter the value in grams (g).
---
#### Steps to Solve the Problem:
1. **Identify the balanced chemical equation for the reaction:**
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
2. **Calculate the moles of HCl and NaOH:**
- Moles of HCl = \(\frac{\text{Given mass of HCl}}{\text{Molar mass of HCl}}\)
- Moles of NaOH = \(\frac{\text{Given mass of NaOH}}{\text{Molar mass of NaOH}}\)
3. **Determine the limiting reactant:**
- Compare mole ratios to find the reactant that will run out first.
4. **Calculate the moles of water (H₂O) produced:**
- Use the mole ratio from the balanced equation to find the moles of H₂O produced by the limiting reactant.
5. **Calculate the mass of water produced:**
- Mass of H₂O = Moles of H₂O \(\times\) Molar mass of H₂O
6. **Round the answer to 2 significant digits.**
---
#### Implementing the Calculation:
1. Moles of HCl:
\[ \frac{32.4\, \text{g}}{36.46\, \text{g/mol}} \approx 0.889\, \text{mol} \]
2. Moles of NaOH:
\[ \frac{26.0\, \text{g}}{40.00\, \text{g/mol}} = 0.650\, \text{mol} \]
3. The balanced equation shows a 1:1 molar ratio between HCl and NaOH. Therefore, NaOH is the limiting reactant because it has fewer](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7c14a699-de9b-4c1d-9356-4bb35f1041ad%2Fa1721a17-b7a4-4eb8-a136-c4e9941bef60%2F8c8mdfm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Chemical Reaction Problem
#### Problem Statement
Aqueous hydrochloric acid (HCl) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H₂O).
Suppose 32.4 g of hydrochloric acid is mixed with 26.0 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits.
#### Instruction
Enter the value in grams (g).
---
#### Steps to Solve the Problem:
1. **Identify the balanced chemical equation for the reaction:**
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
2. **Calculate the moles of HCl and NaOH:**
- Moles of HCl = \(\frac{\text{Given mass of HCl}}{\text{Molar mass of HCl}}\)
- Moles of NaOH = \(\frac{\text{Given mass of NaOH}}{\text{Molar mass of NaOH}}\)
3. **Determine the limiting reactant:**
- Compare mole ratios to find the reactant that will run out first.
4. **Calculate the moles of water (H₂O) produced:**
- Use the mole ratio from the balanced equation to find the moles of H₂O produced by the limiting reactant.
5. **Calculate the mass of water produced:**
- Mass of H₂O = Moles of H₂O \(\times\) Molar mass of H₂O
6. **Round the answer to 2 significant digits.**
---
#### Implementing the Calculation:
1. Moles of HCl:
\[ \frac{32.4\, \text{g}}{36.46\, \text{g/mol}} \approx 0.889\, \text{mol} \]
2. Moles of NaOH:
\[ \frac{26.0\, \text{g}}{40.00\, \text{g/mol}} = 0.650\, \text{mol} \]
3. The balanced equation shows a 1:1 molar ratio between HCl and NaOH. Therefore, NaOH is the limiting reactant because it has fewer
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