Answer :
To determine the percent yield of the reaction between calcium hydroxide [tex]\(\text{Ca(OH)}_2\)[/tex] and phosphoric acid [tex]\(\text{H}_3\text{PO}_4\)[/tex], we need to follow a series of steps involving stoichiometry and yield calculations. Let’s go through it step-by-step:
1. Write down the balanced chemical equation:
[tex]\[ 3 \text{Ca(OH)}_2 + 2 \text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6 \text{H}_2\text{O} \][/tex]
2. Determine the molar masses:
- Calcium Hydroxide, [tex]\(\text{Ca(OH)}_2\)[/tex]:
[tex]\[ \text{Molar mass} = 40.08 \, (\text{Ca}) + 2 \times (16.00 \, (\text{O}) + 1.008 \, (\text{H})) = 74.096 \, \text{g/mol} \][/tex]
- Phosphoric Acid, [tex]\(\text{H}_3\text{PO}_4\)[/tex]:
[tex]\[ \text{Molar mass} = 3 \times 1.008 \, (\text{H}) + 30.974 \, (\text{P}) + 4 \times 16.00 \, (\text{O}) = 97.994 \, \text{g/mol} \][/tex]
- Calcium Phosphate, [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]:
[tex]\[ \text{Molar mass} = 3 \times 40.08 \, (\text{Ca}) + 2 \times 30.974 \, (\text{P}) + 8 \times 16.00 \, (\text{O}) = 310.18 \, \text{g/mol} \][/tex]
3. Calculate the moles of each reactant:
- Moles of [tex]\(\text{Ca(OH)}_2\)[/tex]:
[tex]\[ \text{moles} = \frac{9.8 \, \text{g}}{74.096 \, \text{g/mol}} = 0.1323 \, \text{mol} \][/tex]
- Moles of [tex]\(\text{H}_3\text{PO}_4\)[/tex]:
[tex]\[ \text{moles} = \frac{9.8 \, \text{g}}{97.994 \, \text{g/mol}} = 0.1000 \, \text{mol} \][/tex]
4. Determine the limiting reagent:
The stoichiometric ratio from the balanced equation is 3:2 for [tex]\(\text{Ca(OH)}_2\)[/tex] to [tex]\(\text{H}_3\text{PO}_4\)[/tex].
- For [tex]\(\text{Ca(OH)}_2\)[/tex]:
[tex]\[ \text{moles of } \text{Ca}_3(\text{PO}_4)_2 = \frac{0.1323 \, \text{mol} \, \text{Ca(OH)}_2}{3} = 0.0441 \, \text{mol} \][/tex]
- For [tex]\(\text{H}_3\text{PO}_4\)[/tex]:
[tex]\[ \text{moles of } \text{Ca}_3(\text{PO}_4)_2 = \frac{0.1000 \, \text{mol} \, \text{H}_3\text{PO}_4}{2} = 0.0500 \, \text{mol} \][/tex]
The limiting reagent is [tex]\(\text{Ca(OH)}_2\)[/tex] because it produces fewer moles of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex].
5. Calculate the theoretical yield of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]:
- Moles of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex] from limiting reagent: [tex]\(0.0441 \)[/tex] mol
- Mass of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]:
[tex]\[ \text{mass} = 0.0441 \, \text{mol} \times 310.18 \, \text{g/mol} = 13.67 \, \text{g} \][/tex]
6. Determine the percent yield:
- Actual yield of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]: [tex]\(2.5 \, \text{g}\)[/tex]
- Percent yield:
[tex]\[ \text{percent yield} = \frac{2.5 \, \text{g}}{13.67 \, \text{g}} \times 100\% = 18.29\% \][/tex]
Thus, the percent yield for the reaction is 18.29%.
1. Write down the balanced chemical equation:
[tex]\[ 3 \text{Ca(OH)}_2 + 2 \text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6 \text{H}_2\text{O} \][/tex]
2. Determine the molar masses:
- Calcium Hydroxide, [tex]\(\text{Ca(OH)}_2\)[/tex]:
[tex]\[ \text{Molar mass} = 40.08 \, (\text{Ca}) + 2 \times (16.00 \, (\text{O}) + 1.008 \, (\text{H})) = 74.096 \, \text{g/mol} \][/tex]
- Phosphoric Acid, [tex]\(\text{H}_3\text{PO}_4\)[/tex]:
[tex]\[ \text{Molar mass} = 3 \times 1.008 \, (\text{H}) + 30.974 \, (\text{P}) + 4 \times 16.00 \, (\text{O}) = 97.994 \, \text{g/mol} \][/tex]
- Calcium Phosphate, [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]:
[tex]\[ \text{Molar mass} = 3 \times 40.08 \, (\text{Ca}) + 2 \times 30.974 \, (\text{P}) + 8 \times 16.00 \, (\text{O}) = 310.18 \, \text{g/mol} \][/tex]
3. Calculate the moles of each reactant:
- Moles of [tex]\(\text{Ca(OH)}_2\)[/tex]:
[tex]\[ \text{moles} = \frac{9.8 \, \text{g}}{74.096 \, \text{g/mol}} = 0.1323 \, \text{mol} \][/tex]
- Moles of [tex]\(\text{H}_3\text{PO}_4\)[/tex]:
[tex]\[ \text{moles} = \frac{9.8 \, \text{g}}{97.994 \, \text{g/mol}} = 0.1000 \, \text{mol} \][/tex]
4. Determine the limiting reagent:
The stoichiometric ratio from the balanced equation is 3:2 for [tex]\(\text{Ca(OH)}_2\)[/tex] to [tex]\(\text{H}_3\text{PO}_4\)[/tex].
- For [tex]\(\text{Ca(OH)}_2\)[/tex]:
[tex]\[ \text{moles of } \text{Ca}_3(\text{PO}_4)_2 = \frac{0.1323 \, \text{mol} \, \text{Ca(OH)}_2}{3} = 0.0441 \, \text{mol} \][/tex]
- For [tex]\(\text{H}_3\text{PO}_4\)[/tex]:
[tex]\[ \text{moles of } \text{Ca}_3(\text{PO}_4)_2 = \frac{0.1000 \, \text{mol} \, \text{H}_3\text{PO}_4}{2} = 0.0500 \, \text{mol} \][/tex]
The limiting reagent is [tex]\(\text{Ca(OH)}_2\)[/tex] because it produces fewer moles of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex].
5. Calculate the theoretical yield of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]:
- Moles of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex] from limiting reagent: [tex]\(0.0441 \)[/tex] mol
- Mass of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]:
[tex]\[ \text{mass} = 0.0441 \, \text{mol} \times 310.18 \, \text{g/mol} = 13.67 \, \text{g} \][/tex]
6. Determine the percent yield:
- Actual yield of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]: [tex]\(2.5 \, \text{g}\)[/tex]
- Percent yield:
[tex]\[ \text{percent yield} = \frac{2.5 \, \text{g}}{13.67 \, \text{g}} \times 100\% = 18.29\% \][/tex]
Thus, the percent yield for the reaction is 18.29%.
