What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Na₃PO₄ reacts with 57.0 mL of 0.200 M CrCl₃ in the following chemical reaction? Na₃PO₄(aq) + CrCl₃(aq) → CrPO₄(s) + 3 NaCl(aq)

Answer 1

Limiting reactant problem

When given amounts of both reactants, one needs to find which, if any, is limiting. One way to do this is to divide the moles of each reactant by the corresponding coefficient in the balanced equation.

Na₃PO₄(aq) + CrCl₃(aq) ==> CrPO₄(s) + 3NaCl(aq)

mols Na₃PO₄ = 45.5 ml x 1 L / 1000 ml x 0.300 mol/L = 0.01365 mols (÷1->0.0137)

mols CrCl₃ = 57.0 ml x 1 L / 1000 ml x 0.200 mol/L = 0.0114 mols (÷1->0.0114)

Since 0.0114 is less than 0.0137, the CrCl₃ is the limiting reactant, and moles of CrCl₃ will dictate the amount of product that can be formed.

Mass of precipitate (CrPO₄) = 0.0114 mol CrCl₃ x 1 mol CrPO₄ / mol CrCl₃ x 147 g / mol = 1.68 g CrPO₄