### Part 1: Calculating the Mass of Oxalic Acid Dihydrate (C₂H₂O₄•2H₂O)
1. Convert the Volume of Solution from mL to L:
Given the volume of the solution is 600.0 mL. To convert this to liters (L):
[tex]\[
\text{Volume} = \frac{600.0 \,\text{mL}}{1000} = 0.6 \,\text{L}
\][/tex]
2. Determine the Moles of Oxalic Acid Needed:
The molarity of the oxalic acid solution is given as 3.500 M (moles per liter). Using the volume in liters to find the number of moles:
[tex]\[
\text{Moles of oxalic acid} = \text{Volume} \times \text{Molarity} = 0.6 \,\text{L} \times 3.500 \,\text{M} = 2.1 \,\text{moles}
\][/tex]
3. Calculate the Molar Mass of Oxalic Acid Dihydrate:
- Oxalic acid (C₂H₂O₄) molar mass:
[tex]\[
\text{Molar mass of C}_2\text{H}_2\text{O}_4 = 2 \times 12.01 + 2 \times 1.01 + 4 \times 16.00 = 90.04 \,\text{g/mol}
\][/tex]
- Water (H₂O) molar mass:
[tex]\[
\text{Molar mass of H}_2\text{O} = 2 \times 1.01 + 16.00 = 18.02 \,\text{g/mol}
\][/tex]
- Since there are two water molecules in oxalic acid dihydrate:
[tex]\[
\text{Molar mass of 2H}_2\text{O} = 2 \times 18.02 = 36.04 \,\text{g/mol}
\][/tex]
- Thus, the molar mass of the full oxalic acid dihydrate (C₂H₂O₄•2H₂O) is:
[tex]\[
\text{Molar mass of C}_2\text{H}_2\text{O}_4•2\text{H}_2\text{O} = 90.04 + 36.04 = 126.08 \,\text{g/mol}
\][/tex]
4. Calculate the Mass of Oxalic Acid Dihydrate Required:
Using the moles of oxalic acid calculated and the molar mass of the dihydrate form:
[tex]\[
\text{Mass of oxalic acid dihydrate} = \text{Moles} \times \text{Molar mass} = 2.1 \,\text{moles} \times 126.08 \,\text{g/mol} = 264.768 \,\text{g}
\][/tex]
So, you need 264.768 grams of oxalic acid dihydrate to prepare 600.0 mL of a 3.500 M oxalic acid solution.
### Part 2: Writing the Complete and Net Ionic Equations of the Neutralization Reaction Between NaOH and HCl
1. Balanced Molecular Equation:
[tex]\[
\text{NaOH}(aq) + \text{HCl}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)
\][/tex]
2. Complete Ionic Equation:
Separate all strong electrolytes into their ions:
[tex]\[
\text{Na}^+(aq) + \text{OH}^-(aq) + \text{H}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{Na}^+(aq) + \text{Cl}^-(aq) + \text{H}_2\text{O}(l)
\][/tex]
3. Net Ionic Equation:
Cancel out the spectator ions (Na[tex]\(^+\)[/tex] and Cl[tex]\(^-\)[/tex]) from the complete ionic equation since they appear on both sides:
[tex]\[
\text{OH}^-(aq) + \text{H}^+(aq) \rightarrow \text{H}_2\text{O}(l)
\][/tex]
Thus, the complete ionic equation is:
[tex]\[
\text{Na}^+(aq) + \text{OH}^-(aq) + \text{H}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{Na}^+(aq) + \text{Cl}^-(aq) + \text{H}_2\text{O}(l)
\][/tex]
And the net ionic equation is:
[tex]\[
\text{OH}^-(aq) + \text{H}^+(aq) \rightarrow \text{H}_2\text{O}(l)
\][/tex]
