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?

1 Answer
Meave60
Nov 29, 2015

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"#.

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