Determine the mass of baking soda (84 g/mol) needed to completely react with 531mL of vinegar. Assume vinegar is 5% (m/m) acetic acid (60 g/mol) and the density of vinegar is 1.05 g/mL. O 38 g O 37 g O 40 g O 39 g O 35 g O 36 g

Phlebotomy Essentials
6th Edition
ISBN:9781451194524
Author:Ruth McCall, Cathee M. Tankersley MT(ASCP)
Publisher:Ruth McCall, Cathee M. Tankersley MT(ASCP)
Chapter1: Phlebotomy: Past And Present And The Healthcare Setting
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**Question:**

Determine the mass of baking soda (84 g/mol) needed to completely react with 531 mL of vinegar. Assume vinegar is 5% (m/m) acetic acid (60 g/mol) and the density of vinegar is 1.05 g/mL.

**Options:**

- ○ 38 g
- ○ 37 g
- ○ 40 g
- ○ 39 g
- ○ 35 g
- ○ 36 g

**Explanation:**

To solve this problem, follow these steps:

1. **Calculate the mass of vinegar:**
   \[
   \text{Mass of vinegar} = \text{Volume} \times \text{Density} = 531\, \text{mL} \times 1.05\, \text{g/mL} = 557.55\, \text{g}
   \]

2. **Find the mass of acetic acid in the vinegar:**
   \[
   \text{Mass of acetic acid} = \frac{5}{100} \times 557.55\, \text{g} = 27.8775\, \text{g}
   \]

3. **Calculate moles of acetic acid:**
   \[
   \text{Moles of acetic acid} = \frac{27.8775\, \text{g}}{60\, \text{g/mol}} = 0.464625\, \text{mol}
   \]

4. **Determine moles of baking soda needed:**
   The reaction between baking soda (NaHCO₃) and acetic acid (CH₃COOH) is:
   \[
   \text{NaHCO}_3 + \text{CH}_3\text{COOH} \rightarrow \text{CO}_2 + \text{H}_2\text{O} + \text{NaCH}_3\text{COO}
   \]
   This is a 1:1 molar reaction, so moles of baking soda needed = moles of acetic acid.

5. **Calculate the mass of baking soda required:**
   \[
   \text{Mass of baking soda} = 0.464625\, \text{mol} \times 84\, \text{g/mol} = 39.0245\, \text{g}
   \]
Transcribed Image Text:**Question:** Determine the mass of baking soda (84 g/mol) needed to completely react with 531 mL of vinegar. Assume vinegar is 5% (m/m) acetic acid (60 g/mol) and the density of vinegar is 1.05 g/mL. **Options:** - ○ 38 g - ○ 37 g - ○ 40 g - ○ 39 g - ○ 35 g - ○ 36 g **Explanation:** To solve this problem, follow these steps: 1. **Calculate the mass of vinegar:** \[ \text{Mass of vinegar} = \text{Volume} \times \text{Density} = 531\, \text{mL} \times 1.05\, \text{g/mL} = 557.55\, \text{g} \] 2. **Find the mass of acetic acid in the vinegar:** \[ \text{Mass of acetic acid} = \frac{5}{100} \times 557.55\, \text{g} = 27.8775\, \text{g} \] 3. **Calculate moles of acetic acid:** \[ \text{Moles of acetic acid} = \frac{27.8775\, \text{g}}{60\, \text{g/mol}} = 0.464625\, \text{mol} \] 4. **Determine moles of baking soda needed:** The reaction between baking soda (NaHCO₃) and acetic acid (CH₃COOH) is: \[ \text{NaHCO}_3 + \text{CH}_3\text{COOH} \rightarrow \text{CO}_2 + \text{H}_2\text{O} + \text{NaCH}_3\text{COO} \] This is a 1:1 molar reaction, so moles of baking soda needed = moles of acetic acid. 5. **Calculate the mass of baking soda required:** \[ \text{Mass of baking soda} = 0.464625\, \text{mol} \times 84\, \text{g/mol} = 39.0245\, \text{g} \]
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