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

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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
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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