If 0.5 mole of BaCl2 is mixed with 0.2 moles of Na3PO4 calculate the maximum no of Mole of Ba(PO4)2 that will be formed?

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

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 ratio from the balanced equation with barium phosphate in the numerator to get moles of barium phosphate.

##0.5"mol BaCl"_2xx(1"mol Ba"_3"(PO"_4")"_2)/(3"mol BaCl"_2)="0.2 mol Ba"_3"(PO"_4")"_2"## (rounded to one significant figure)

##0.2"mol Na"_3"PO"_4xx(1"mol Ba"_3"(PO"_4")"_2)/(2"mol Na"_3"PO"_4)="0.1 mol Ba"_3"(PO"_4")"_2"##

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