To determine how many moles of [tex]\( \text{CaCl}_2 \)[/tex] remain if 0.10 mol of [tex]\( \text{Na}_3\text{PO}_4 \)[/tex] reacts with 0.40 mol of [tex]\( \text{CaCl}_2 \)[/tex], follow these steps:
1. Write down the balanced chemical equation:
[tex]\[ 2 \text{Na}_3\text{PO}_4(aq) + 3 \text{CaCl}_2(aq) \rightarrow \text{Ca}_3(\text{PO}_4)_2(s) + 6 \text{NaCl}(aq) \][/tex]
2. Identify the stoichiometric coefficients:
- For [tex]\( \text{Na}_3\text{PO}_4 \)[/tex]: 2
- For [tex]\( \text{CaCl}_2 \)[/tex]: 3
3. Convert the stoichiometric coefficients to a mole ratio:
The ratio of [tex]\( \text{CaCl}_2 \)[/tex] to [tex]\( \text{Na}_3\text{PO}_4 \)[/tex] is:
[tex]\[ \frac{3 \text{ moles of } \text{CaCl}_2}{2 \text{ moles of } \text{Na}_3\text{PO}_4} = 1.5 \][/tex]
4. Calculate the amount of [tex]\( \text{CaCl}_2 \)[/tex] needed to react with 0.10 mol of [tex]\( \text{Na}_3\text{PO}_4 \)[/tex]:
Since the ratio is 1.5, the moles of [tex]\( \text{CaCl}_2 \)[/tex] required are:
[tex]\[ 0.10 \text{ mol of } \text{Na}_3\text{PO}_4 \times 1.5 = 0.15 \text{ mol of } \text{CaCl}_2 \][/tex]
5. Determine the remaining amount of [tex]\( \text{CaCl}_2 \)[/tex]:
We started with 0.40 mol of [tex]\( \text{CaCl}_2 \)[/tex] and 0.15 mol is used up in the reaction.
[tex]\[ 0.40 \text{ mol of } \text{CaCl}_2 - 0.15 \text{ mol of } \text{CaCl}_2 = 0.25 \text{ mol of } \text{CaCl}_2 \][/tex]
6. Conclusion:
The amount of [tex]\( \text{CaCl}_2 \)[/tex] remaining is:
[tex]\[ \boxed{0.25 \text{ mol}} \][/tex]
Therefore, the correct answer is:
A. [tex]\( 0.25 \text{ mol} \)[/tex]
