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?

Question

Henry Antrim · Accepted Answer

The maximum amount of #"Ba"_3"(PO"_4")"_2"# that can be produced is #"0.1 mol"#.

This question concerns the limiting reactant (reagent), which establishes the maximum amount of a product that can be produced. The limiting reactant is the one that produces the least amount of barium phosphate. Create a balanced equation first. #"3BaCl"_2("aq")"+ 2Na"_3"PO"_4("aq")##rarr##"Ba"_3"(PO"_4)_2("s") + "6NaCl(aq)"# To get moles of barium phosphate, multiply the moles of each reactant by the mole ratio found in the balanced equation with barium phosphate in the numerator. #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"#.

Owen Ambrose · Answer

undefined To determine the maximum amount of Ba3(PO4)2 that can be produced, you need to first identify the limiting reactant. Then, you can use stoichiometry to calculate the maximum amount of product formed based on the amount of limiting reactant.