Report Form Experiment 6 Stoichiometry and Limiting Reactant

Report Form, Experiment 6 Stoichiometry and Limiting Reactants Name Jsmin Suaren Data and Calculations: The preliminary calculations must be completed and approved by the instructor before Lab the experimental work is begun. I. Balance the equation for the reaction and write the coefficients in the blanks for the total molecular equation and the net ionic equation. 2. Determine the following quantities for your experiment: The number of grams in one mole NasPO, 12H,O The number of grams in one mole of BaCl2 2H20 The number of grams in one mole o The number of grams in one mole of NaCI grams 319.94 244.2 3 grams f Bas(PO4h () grams 5,44 3. Mole Assignment: (Assigned by Lab Instructor) Compound Assigned number of moles Assigned number of grams 00029 2.40 BaCl 2H,O The limiting reagent is: The theoretical number of moles of Bas(PO4)2 (6) produced is The theoretical number of grams of Ba(PO4h (s produced is First Attempt 00250 -Second Attempt : Instructor approval. 57

Solution

The balanced chemical equation is

2 Na₃PO₄.12H₂O (aq) + 3 BaCl₂.2H₂O (aq) ---------> Ba₃(PO₄)₂ (s) + 6 NaCl (aq) + 30 H₂O (l)

As per the balanced stoichiometric equation,

2 moles Na₃PO₄.12H₂O = 3 moles BaCl₂.2H₂O = 1 mole Ba₃(PO₄)₂

Determine the limiting reactant.

0.00625 mole Na₃PO₄.12H₂O = (0.00625 mole Na₃PO₄.12H₂O)*(3 moles BaCl₂.2H₂O/2 moles Na₃PO₄.12H₂O) = 0.009375 mole BaCl₂.2H₂O

0.00250 mole BaCl₂.2H₂O = (0.00250 mole BaCl₂.2H₂O)*(2 moles Na₃PO₄.12H₂O/3 moles BaCl₂.2H₂O) = 0.001667 mole Na₃PO₄.12H₂O

Offcourse, we do not have 0.009375 mole BaCl₂.2H₂O, but we have excess Na₃PO₄.12H₂O than required to react with 0.00250 mole BaCl₂.2H₂O.

Therefore, the limiting reactant is BaCl₂.2H₂O (ans).

The yield of the product is determined by the limiting reactant.

Theoretical number of moles of Ba₃PO₄ possible = (0.00250 mole BaCl₂.2H₂O)*(1 mole Ba₃(PO₄)₂/3 moles BaCl₂.2H₂O) = 0.000833 mole Ba₃(PO₄)₂ (ans).

Molar mass of Ba₃(PO₄)₂ = 601.93 g/mol; therefore, theoretical yield of Ba₃(PO₄)₂ = (0.000833 mole)*(601.93 g/mol) = 0.5014 g (ans).