A reaction of 44.7 g of Na and 55.1 g of Br2 yields 65.9 g of NaBr . What is the percent yield? 2Na(s)+Br2(g)⟶2NaBr(s)

Question

A reaction of 44.7 g of Na and 55.1 g of Br2 yields 65.9 g of NaBr. What is the percent yield?

2Na(s)+Br2(g)⟶2NaBr(s)

percent yield:____?______%

Answer 1

Before one can find the % yield, one must first find the theoretical yield. Then, % yield = actual yield / theoretical yield (x100%).

2Na(s) + Br₂(g) ==> 2NaBr(s) ... balanced equation

One easy way to find the limiting reactant is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever result is less is the limiting reactant.

For Na: 44.7 g Na x 1 mol Na / 23 g = 1.94 moles Na (÷2->0.97)

For Br₂: 55.1 g Br₂ x 1 mol Br₂ / 160 g = 0.344 moles Br₂ (÷1->0.34)

Since 0.34 is less than 0.97, Br₂ is LIMITING

Now use the MOLES of the limiting reactant to find the theoretical yield of NaBr:

0.344 mols Br₂ x 2 mols NaBr / 1 mol Br₂ x 103 g NaBr / mol = 70.9 g NaBr = theoretical yield

% yield = 65.9 g / 70.9 g (x100%) = 92.9% yield