To solve the problem, we will follow these steps:
### Step 1: Write the balanced chemical equation.
The balanced chemical equation for the reaction is:
\[ 3 \text{CaCO}_3 + 2 \text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 3 \text{H}_2\text{O} + 3 \text{CO}_2 \]
### Step 2: Calculate the moles of \( \text{CaCO}_3 \).
Given mass of \( \text{CaCO}_3 = 50 \, \text{g} \) and its molar mass is \( 100 \, \text{g/mol} \):
\[
\text{Moles of } \text{CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{50 \, \text{g}}{100 \, \text{g/mol}} = 0.5 \, \text{mol}
\]
### Step 3: Calculate the moles of \( \text{H}_3\text{PO}_4 \).
Given mass of \( \text{H}_3\text{PO}_4 = 68.6 \, \text{g} \) and its molar mass is \( 98 \, \text{g/mol} \):
\[
\text{Moles of } \text{H}_3\text{PO}_4 = \frac{68.6 \, \text{g}}{98 \, \text{g/mol}} \approx 0.7 \, \text{mol}
\]
### Step 4: Determine the limiting reagent.
From the balanced equation, \( 3 \) moles of \( \text{CaCO}_3 \) react with \( 2 \) moles of \( \text{H}_3\text{PO}_4 \). Therefore, \( 1 \) mole of \( \text{CaCO}_3 \) requires \( \frac{2}{3} \) moles of \( \text{H}_3\text{PO}_4 \).
For \( 0.5 \) moles of \( \text{CaCO}_3 \):
\[
\text{Moles of } \text{H}_3\text{PO}_4 \text{ required} = 0.5 \times \frac{2}{3} = 0.333 \, \text{mol}
\]
Since we have \( 0.7 \, \text{mol} \) of \( \text{H}_3\text{PO}_4 \) available, \( \text{CaCO}_3 \) is the limiting reagent.
### Step 5: Calculate the moles of product formed.
From the balanced equation, \( 3 \) moles of \( \text{CaCO}_3 \) produce \( 1 \) mole of \( \text{Ca}_3(\text{PO}_4)_2 \). Thus, \( 0.5 \) moles of \( \text{CaCO}_3 \) will produce:
\[
\text{Moles of } \text{Ca}_3(\text{PO}_4)_2 = \frac{0.5}{3} \approx 0.167 \, \text{mol}
\]
### Step 6: Calculate the mass of \( \text{Ca}_3(\text{PO}_4)_2 \) produced.
The molar mass of \( \text{Ca}_3(\text{PO}_4)_2 \) is \( 310 \, \text{g/mol} \):
\[
\text{Mass of } \text{Ca}_3(\text{PO}_4)_2 = 0.167 \, \text{mol} \times 310 \, \text{g/mol} \approx 51.67 \, \text{g}
\]
### Step 7: Calculate the moles of \( \text{CO}_2 \) produced.
From the balanced equation, \( 3 \) moles of \( \text{CaCO}_3 \) produce \( 3 \) moles of \( \text{CO}_2 \). Thus, \( 0.5 \) moles of \( \text{CaCO}_3 \) will produce:
\[
\text{Moles of } \text{CO}_2 = 0.5 \, \text{mol}
\]
### Step 8: Calculate the remaining moles of \( \text{H}_3\text{PO}_4 \).
Total moles of \( \text{H}_3\text{PO}_4 \) = \( 0.7 \, \text{mol} \)
Moles used = \( 0.333 \, \text{mol} \)
Remaining moles = \( 0.7 - 0.333 = 0.367 \, \text{mol} \)
### Step 9: Calculate the mass of unreacted \( \text{H}_3\text{PO}_4 \).
\[
\text{Mass of unreacted } \text{H}_3\text{PO}_4 = 0.367 \, \text{mol} \times 98 \, \text{g/mol} \approx 35.96 \, \text{g}
\]
### Summary of Results:
- Mass of \( \text{Ca}_3(\text{PO}_4)_2 \) produced: \( 51.67 \, \text{g} \)
- Moles of \( \text{CO}_2 \) produced: \( 0.5 \, \text{mol} \)
- Mass of unreacted \( \text{H}_3\text{PO}_4 \): \( 35.96 \, \text{g} \)
### Correct Options:
- A: Mass of \( \text{Ca}_3(\text{PO}_4)_2 \) produced is correct.
- B: Mass of unreacted \( \text{H}_3\text{PO}_4 \) is correct.
- C: Moles of \( \text{CO}_2 \) produced is correct.
