Question

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?

Answer 1

The maximum amount of `"Ba"_3"(PO"_4")"_2"` that can be produced is `"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.

Start with a balanced equation.

`"3BaCl"_2("aq")"+ 2Na"_3"PO"_4("aq")` `rarr` `"Ba"_3"(PO"_4)_2("s") + "6NaCl(aq)"`

Multiply the moles of each reactant times the mole ratio from the balanced equation with barium phosphate in the numerator to get moles of barium phosphate.

`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"`

`"Na"_3"PO"_4"` is the limiting reactant, so the maximum amount of `"Ba"_3"(PO"_4")"_2"` that can be produced by this reaction is `"0.1 mol"`.