Theoretical yield

Question

Consider the following chemical reaction:

Na2Cr2O7(aq) + BaCl2(aq) = NaCl(aq) + BaCr2O7(s)

2.20 g sodium dichromate is mixed with 0.0150 moles barium chloride. Calculate

theoretical vield of barium dichromate

Na = 22.99 g/mole, Cr = 52.00 g/mole, O= 16.00 g/mole, Ba = 137.3 g/mole, CI=35.45 g/mole

Answer 1

Na₂Cr₂O₇(aq) + BaCl₂(aq) ==> NaCl(aq) + BaCr₂O₇(s) ... unbalanced equation

First, balance the equation:

Na₂Cr₂O₇(aq) + BaCl₂(aq) ==> 2NaCl(aq) + BaCr₂O₇(s) ... balanced equation

Next, find the limiting reactant. One easy way is to divided the moles of each reactant by the corresponding coefficient in the balanced equation and whichever is less represents the limiting reactant.

For Na₂Cr₂O₇: 2.20 g Na₂Cr₂O₇ x 1 mol / 261.97 g = 0.008398 mols (÷1->0.08398)

For BaCl₂: 0.0150 mols (÷1->0.015)

BaCl₂ is the limiting reactant since 0.015 is less than 0.084.

Now use the MOLE of the limiting reactant (BaCl₂) to find the theoretical yield of BaCr₂O₇(s)

0.0150 mols BaCl₂ x 1 mol BaCr₂O₇(s) / mol BaCl₂ x 353.3 g BaCr₂O₇(s)/mol = 5.30 g BaCr₂O₇(s)