If "0.5 mol BaCl"_2" reacts with "0.2 mol Na"_3"PO"_4", what is the maximum amount of "Ba"_3"(PO"_4)_2" that can be produced?

Nov 29, 2015

The maximum amount of $\text{Ba"_3"(PO"_4")"_2}$ that can be produced is $\text{0.1 mol}$.

Explanation:

This is a limiting reactant (reagent) question. The reactant that produces the least amount of barium phosphate is the limiting reactant. The limiting reactant determines the maximum amount of a product can be produced.

"3BaCl"_2("aq")"+ 2Na"_3"PO"_4("aq")$\rightarrow$$\text{Ba"_3"(PO"_4)_2("s") + "6NaCl(aq)}$
0.5color(red)cancel(color(black)("mol BaCl"_2))xx(1"mol Ba"_3"(PO"_4")"_2)/(3color(red)cancel(color(black)("mol BaCl"_2)))="0.2 mol Ba"_3"(PO"_4")"_2" (rounded to one significant figure)
0.2color(red)cancel(color(black)("mol Na"_3"PO"_4))xx(1"mol Ba"_3"(PO"_4")"_2)/(2color(red)cancel(color(black)("mol Na"_3"PO"_4)))="0.1 mol Ba"_3"(PO"_4")"_2"
$\text{Na"_3"PO"_4}$ is the limiting reactant, so the maximum amount of $\text{Ba"_3"(PO"_4")"_2}$ that can be produced by this reaction is $\text{0.1 mol}$.