[References] What volume of 0.125 M oxalic acid, H₂C2O4, is required to react with 92.9 mL of 0.372 M NaOH? H₂C₂O4 (aq) + 2 NaOH(aq) → Na₂C₂O4 (aq) + 2 H₂O(l) Volume = Submit Answer L Try Another Version 8 item attempts remaining

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**Stoichiometry Problem: Determining Volume of Oxalic Acid for Neutralization Reaction**

**Question:**
What volume of 0.125 M oxalic acid, \( \mathrm{H_2C_2O_4} \), is required to react with 92.9 mL of 0.372 M \( \mathrm{NaOH} \)?

**Chemical Equation:**
\[ \mathrm{H_2C_2O_4(aq) + 2 \ NaOH(aq) \rightarrow Na_2C_2O_4(aq) + 2 \ H_2O(\ell)} \]

**Answer:**
Volume = ______ L

**Solution Steps:**

1. **Write the balanced chemical equation:**
   \[ \mathrm{H_2C_2O_4(aq) + 2 \ NaOH(aq) \rightarrow Na_2C_2O_4(aq) + 2 \ H_2O(\ell)} \]

2. **Identify known quantities:**
   - Concentration of \( \mathrm{NaOH} = 0.372 \ M \)
   - Volume of \( \mathrm{NaOH} = 92.9 \ mL \) (convert to liters by dividing by 1000: 92.9 mL = 0.0929 L)
   - Concentration of \( \mathrm{H_2C_2O_4} = 0.125 \ M \)

3. **Calculate moles of \( \mathrm{NaOH} \):**
   \[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} \]
   \[ \text{Moles of NaOH} = 0.372 \ \text{M} \times 0.0929 \ \text{L} \]
   \[ \text{Moles of NaOH} = 0.0345348 \ \text{moles} \]

4. **Use stoichiometry to find moles of \( \mathrm{H_2C_2O_4} \):**
   From the balanced equation, 1 mole of \( \mathrm{H_2C_2O_4} \) reacts with 2 moles of \( \mathrm{NaOH} \).
   \[ \
Transcribed Image Text:**Stoichiometry Problem: Determining Volume of Oxalic Acid for Neutralization Reaction** **Question:** What volume of 0.125 M oxalic acid, \( \mathrm{H_2C_2O_4} \), is required to react with 92.9 mL of 0.372 M \( \mathrm{NaOH} \)? **Chemical Equation:** \[ \mathrm{H_2C_2O_4(aq) + 2 \ NaOH(aq) \rightarrow Na_2C_2O_4(aq) + 2 \ H_2O(\ell)} \] **Answer:** Volume = ______ L **Solution Steps:** 1. **Write the balanced chemical equation:** \[ \mathrm{H_2C_2O_4(aq) + 2 \ NaOH(aq) \rightarrow Na_2C_2O_4(aq) + 2 \ H_2O(\ell)} \] 2. **Identify known quantities:** - Concentration of \( \mathrm{NaOH} = 0.372 \ M \) - Volume of \( \mathrm{NaOH} = 92.9 \ mL \) (convert to liters by dividing by 1000: 92.9 mL = 0.0929 L) - Concentration of \( \mathrm{H_2C_2O_4} = 0.125 \ M \) 3. **Calculate moles of \( \mathrm{NaOH} \):** \[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} \] \[ \text{Moles of NaOH} = 0.372 \ \text{M} \times 0.0929 \ \text{L} \] \[ \text{Moles of NaOH} = 0.0345348 \ \text{moles} \] 4. **Use stoichiometry to find moles of \( \mathrm{H_2C_2O_4} \):** From the balanced equation, 1 mole of \( \mathrm{H_2C_2O_4} \) reacts with 2 moles of \( \mathrm{NaOH} \). \[ \
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