To determine the maximum amount of $ \text{Ba}_3(\text{PO}_4)_2 $ that can be formed when $0.50 \, \text{mol}$ of $ \text{BaCl}_2 $ is mixed with $0.20 \, \text{mol}$ of $ \text{Na}_3\text{PO}_4 $, we need to look at the stoichiometry of the reaction between barium chloride and sodium phosphate. The balanced chemical equation for the formation of barium phosphate and sodium chloride is:

$$ \text{3BaCl}_2 + 2\text{Na}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6\text{NaCl} $$

This equation tells us that 3 moles of barium chloride react with 2 moles of sodium phosphate to produce 1 mole of barium phosphate. From this, we can calculate the mole ratio needed for the complete reaction:

• Ratio for $ \text{BaCl}_2 : \text{Ba}_3(\text{PO}_4)_2 = 3:1 $


• Ratio for $ \text{Na}_3\text{PO}_4 : \text{Ba}_3(\text{PO}_4)_2 = 2:1 $

Given amounts are $0.50 \, \text{mol}$ of $ \text{BaCl}_2 $ and $0.20 \, \text{mol}$ of $ \text{Na}_3\text{PO}_4 $. To find out which reactant is the limiting reagent, we calculate how much product each could produce if it were completely consumed:

• For $ \text{BaCl}_2 $:

$$ \frac{\text{Moles of BaCl}_2}{\text{Mole ratio for BaCl}_2} = \frac{0.50 \, \text{mol}}{3} = 0.167 \, \text{mols of } \text{Ba}_3(\text{PO}_4)_2 $$

• For $ \text{Na}_3\text{PO}_4 $:

$$ \frac{\text{Moles of Na}_3\text{PO}_4}{\text{Mole ratio for Na}_3\text{PO}_4} = \frac{0.20 \, \text{mol}}{2} = 0.1 \, \text{mols of } \text{Ba}_3(\text{PO}_4)_2 $$

From the calculations, $ \text{Na}_3\text{PO}_4 $ yields less $ \text{Ba}_3(\text{PO}_4)_2 $ than $ \text{BaCl}_2 $ does; therefore, $ \text{Na}_3\text{PO}_4 $ is the limiting reactant. This means the maximum amount of $ \text{Ba}_3(\text{PO}_4)_2 $ that can be formed from the given amounts of reactants is $0.10 \, \text{mol}$.

So, the answer is Option D: 0.10.