100.0 mL of a 0.250 mol/L solution of barium nitrate reacts with 150.0 mL of a 0.125 mol/L solution of sodium phosphate. What is expected mass of the precipitate that is formed?

Answer 1

First, write the correctly balanced equation:

3Ba(NO₃)₂(aq) + 2Na₃PO₄(aq) ==> Ba₃(PO₄)₂(s) + 6NaNO₃(aq) ... balanced equation

Next, find moles of each reactant to see if either is limiting:

moles Ba(NO₃)₂ = 0.100 L x 0.250 mol/L = 0.0250 moles Ba(NO₃)₂

moles Na₃PO₄= 0.150 L x 0.125 mol/L = 0.01875 moles Na₃PO₄

Because it takes 3 moles Ba(NO₃)₂ for each 2 moles Na₃PO₄, the Ba(NO₃)₂ is LIMITING

Finally, using the limiting reactant, the stoichiometry of the balanced equation, and dimensional analysis, we find the mass of precipitate Ba₃(PO₄)₂

0.0250 mol Ba(NO₃)₂ x 1 mol Ba₃(PO₄)₂ / 3 mol Ba(NO₃)₂ x 602 g Ba₃(PO₄)₂ / mol = 5.02 g Ba₃(PO₄)₂