Answer :
To determine the limiting reactant in the reaction between calcium phosphate and sulfuric acid, we need to compare the amount of each reactant available with the stoichiometric amounts required. Here's a detailed step-by-step solution:
1. Identify the Balanced Chemical Equation:
[tex]\[
\text{Ca}_3(\text{PO}_4)_2 + 3 \text{H}_2\text{SO}_4 \rightarrow 3 \text{CaSO}_4 + 2 \text{H}_3\text{PO}_4
\][/tex]
This equation tells us that 1 mole of calcium phosphate reacts with 3 moles of sulfuric acid.
2. Calculate Molar Masses:
- The molar mass of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex] is approximately 310.18 g/mol.
- The molar mass of [tex]\(\text{H}_2\text{SO}_4\)[/tex] is approximately 98.08 g/mol.
3. Calculate Moles of Reactants:
- Moles of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]:
[tex]\[
\frac{129 \, \text{g}}{310.18 \, \text{g/mol}} \approx 0.416 \, \text{mol}
\][/tex]
- Moles of [tex]\(\text{H}_2\text{SO}_4\)[/tex]:
[tex]\[
\frac{97.4 \, \text{g}}{98.08 \, \text{g/mol}} \approx 0.993 \, \text{mol}
\][/tex]
4. Compare Stoichiometric Ratios:
- According to the reaction, 1 mole of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex] requires 3 moles of [tex]\(\text{H}_2\text{SO}_4\)[/tex].
- Calculate the ratio of available moles to the required moles:
- [tex]\(\text{For } \text{Ca}_3(\text{PO}_4)_2\)[/tex], we have 0.416 moles, needing 3 times more moles of [tex]\(\text{H}_2\text{SO}_4\)[/tex].
- [tex]\(\text{For } \text{H}_2\text{SO}_4\)[/tex], divide the moles by 3 (because 3 moles are required per mole of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]):
[tex]\[
\frac{0.993}{3} \approx 0.331 \, \text{mol equivalent to 1 mol of } \text{Ca}_3(\text{PO}_4)_2
\][/tex]
5. Determine the Limiting Reactant:
- Compare the calculated equivalent moles:
- We can only "support" 0.331 mol of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex] with the available [tex]\(\text{H}_2\text{SO}_4\)[/tex], which is less than the 0.416 mol we have.
- Therefore, [tex]\(\text{H}_2\text{SO}_4\)[/tex] is insufficient to react with all of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex], making [tex]\( \text{H}_2\text{SO}_4 \)[/tex] the limiting reactant.
The limiting reactant is [tex]\( \text{H}_2\text{SO}_4 \)[/tex] (option B).
1. Identify the Balanced Chemical Equation:
[tex]\[
\text{Ca}_3(\text{PO}_4)_2 + 3 \text{H}_2\text{SO}_4 \rightarrow 3 \text{CaSO}_4 + 2 \text{H}_3\text{PO}_4
\][/tex]
This equation tells us that 1 mole of calcium phosphate reacts with 3 moles of sulfuric acid.
2. Calculate Molar Masses:
- The molar mass of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex] is approximately 310.18 g/mol.
- The molar mass of [tex]\(\text{H}_2\text{SO}_4\)[/tex] is approximately 98.08 g/mol.
3. Calculate Moles of Reactants:
- Moles of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]:
[tex]\[
\frac{129 \, \text{g}}{310.18 \, \text{g/mol}} \approx 0.416 \, \text{mol}
\][/tex]
- Moles of [tex]\(\text{H}_2\text{SO}_4\)[/tex]:
[tex]\[
\frac{97.4 \, \text{g}}{98.08 \, \text{g/mol}} \approx 0.993 \, \text{mol}
\][/tex]
4. Compare Stoichiometric Ratios:
- According to the reaction, 1 mole of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex] requires 3 moles of [tex]\(\text{H}_2\text{SO}_4\)[/tex].
- Calculate the ratio of available moles to the required moles:
- [tex]\(\text{For } \text{Ca}_3(\text{PO}_4)_2\)[/tex], we have 0.416 moles, needing 3 times more moles of [tex]\(\text{H}_2\text{SO}_4\)[/tex].
- [tex]\(\text{For } \text{H}_2\text{SO}_4\)[/tex], divide the moles by 3 (because 3 moles are required per mole of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]):
[tex]\[
\frac{0.993}{3} \approx 0.331 \, \text{mol equivalent to 1 mol of } \text{Ca}_3(\text{PO}_4)_2
\][/tex]
5. Determine the Limiting Reactant:
- Compare the calculated equivalent moles:
- We can only "support" 0.331 mol of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex] with the available [tex]\(\text{H}_2\text{SO}_4\)[/tex], which is less than the 0.416 mol we have.
- Therefore, [tex]\(\text{H}_2\text{SO}_4\)[/tex] is insufficient to react with all of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex], making [tex]\( \text{H}_2\text{SO}_4 \)[/tex] the limiting reactant.
The limiting reactant is [tex]\( \text{H}_2\text{SO}_4 \)[/tex] (option B).
