`100 ml "of" 0.06 M Ca (NO_(3))_(2)` is added to `50 mL` of `0.06 M Na_(2)C_(2) O_(4)`. After the reaction is complete.

Options

A. 0.003 moles of calcium oxalate will get precipated
B. 0.03 M of excess `Ca^(2+)` will remains in excess.
C. `Na_(2)C_(2)O_(4)` is the limiting reagent
D. `Ca(NO_(3))_(2)` is the excess reagent.

Explanation

To solve the problem step by step, we will analyze the reaction between calcium nitrate and sodium oxalate, calculate the millimoles of each reactant, identify the limiting reagent, and determine the concentration of calcium ions left after the reaction.

### Step 1: Write the balanced chemical equation
The reaction between calcium nitrate \((\text{Ca(NO}_3)_2)\) and sodium oxalate \((\text{Na}_2\text{C}_2\text{O}_4)\) can be represented as:

\[ \text{Ca(NO}_3)_2 + \text{Na}_2\text{C}_2\text{O}_4 \rightarrow \text{CaC}_2\text{O}_4 + 2 \text{NaNO}_3 \]

### Step 2: Calculate the millimoles of each reactant
1. **For calcium nitrate \((\text{Ca(NO}_3)_2)\)**:
- Volume = 100 mL = 0.1 L
- Molarity = 0.06 M
- Millimoles = Molarity × Volume (in L)
\[ \text{Millimoles of Ca(NO}_3)_2 = 0.06 \, \text{mol/L} \times 0.1 \, \text{L} = 0.006 \, \text{mol} = 6 \, \text{mmol} \]

2. **For sodium oxalate \((\text{Na}_2\text{C}_2\text{O}_4)\)**:
- Volume = 50 mL = 0.05 L
- Molarity = 0.06 M
- Millimoles = Molarity × Volume (in L)
\[ \text{Millimoles of Na}_2\text{C}_2\text{O}_4 = 0.06 \, \text{mol/L} \times 0.05 \, \text{L} = 0.003 \, \text{mol} = 3 \, \text{mmol} \]

### Step 3: Identify the limiting reagent
- From the balanced equation, we see that 1 mole of \(\text{Ca(NO}_3)_2\) reacts with 1 mole of \(\text{Na}_2\text{C}_2\text{O}_4\).
- We have 6 mmol of \(\text{Ca(NO}_3)_2\) and 3 mmol of \(\text{Na}_2\text{C}_2\text{O}_4\).
- Since \(\text{Na}_2\text{C}_2\text{O}_4\) has fewer millimoles, it is the limiting reagent.

### Step 4: Calculate the excess of calcium nitrate
- The amount of \(\text{Ca(NO}_3)_2\) that reacts with 3 mmol of \(\text{Na}_2\text{C}_2\text{O}_4\) is also 3 mmol.
- Therefore, the excess of \(\text{Ca(NO}_3)_2\) is:
\[ \text{Excess of Ca(NO}_3)_2 = 6 \, \text{mmol} - 3 \, \text{mmol} = 3 \, \text{mmol} \]

### Step 5: Calculate the concentration of \(\text{Ca}^{2+}\) ions
- After the reaction, the amount of \(\text{Ca}^{2+}\) ions remaining in the solution is equal to the excess \(\text{Ca(NO}_3)_2\) that did not react.
- The total volume of the solution after mixing is:
\[ \text{Total volume} = 100 \, \text{mL} + 50 \, \text{mL} = 150 \, \text{mL} = 0.150 \, \text{L} \]
- The concentration of \(\text{Ca}^{2+}\) ions is:
\[ \text{Concentration of Ca}^{2+} = \frac{\text{millimoles of Ca}^{2+}}{\text{total volume in L}} = \frac{3 \, \text{mmol}}{0.150 \, \text{L}} = 0.02 \, \text{M} \]

### Summary of Results
- Limiting reagent: \(\text{Na}_2\text{C}_2\text{O}_4\)
- Excess reagent: \(\text{Ca(NO}_3)_2\)
- Concentration of \(\text{Ca}^{2+}\) ions: \(0.02 \, \text{M}\)