To solve this problem, we start by identifying the chemical reaction and the information given. The balanced chemical equation for the reaction is:
[tex]\[
\text{Li}_3\text{PO}_4(aq) + 3\text{CaCl}_2(aq) \longrightarrow \text{Ca}_3(\text{PO}_4)_2(s) + 3\text{LiCl}(aq)
\][/tex]
Given data:
- Mass of lithium phosphate ([tex]\(\text{Li}_3\text{PO}_4\)[/tex]): 16.466 grams
We need to find the mass of calcium phosphate ([tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]) produced.
### Step-by-Step Solution:
1. Calculate the molar masses of the reactants and products
- Molar mass of [tex]\(\text{Li}_3\text{PO}_4\)[/tex]:
[tex]\[
3 (\text{Li}) + 1 (\text{P}) + 4 (\text{O}) = 3 (6.94) + 30.97 + 4 (16.00) = 20.82 + 30.97 + 64.00 = 115.79 \text{ g/mol}
\][/tex]
- Molar mass of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]:
[tex]\[
3 (\text{Ca}) + 2 (\text{P}) + 8 (\text{O}) = 3 (40.08) + 2 (30.97) + 8 (16.00) = 120.24 + 61.94 + 128.00 = 310.18 \text{ g/mol}
\][/tex]
2. Calculate the number of moles of [tex]\(\text{Li}_3\text{PO}_4\)[/tex]
Using the mass and the molar mass:
[tex]\[
\text{Moles of } \text{Li}_3\text{PO}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{16.466 \text{ g}}{115.79 \text{ g/mol}} \approx 0.1422 \text{ mol}
\][/tex]
3. Determine the moles of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex] produced
According to the balanced chemical equation, 1 mole of [tex]\(\text{Li}_3\text{PO}_4\)[/tex] produces 1 mole of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]. So, the moles of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex] produced will be equal to the moles of [tex]\(\text{Li}_3\text{PO}_4\)[/tex] given:
[tex]\[
\text{Moles of } \text{Ca}_3(\text{PO}_4)_2 = 0.1422 \text{ mol}
\][/tex]
4. Calculate the mass of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]
Using the moles and molar mass of [tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]:
[tex]\[
\text{Mass of } \text{Ca}_3(\text{PO}_4)_2 = \text{moles} \times \text{molar mass} = 0.1422 \text{ mol} \times 310.18 \text{ g/mol} \approx 44.09 \text{ g}
\][/tex]
Thus, the mass of calcium phosphate ([tex]\(\text{Ca}_3(\text{PO}_4)_2\)[/tex]) that can be made from reacting 16.466 grams of lithium phosphate ([tex]\(\text{Li}_3\text{PO}_4\)[/tex]) is approximately 44.09 grams.
