Chemistry reaction

Question

A reaction of 52.3 g of Na and 36.5 g of Br 2 yields 40.5 g of NaBr. What is the percent yield? 2Na(s)+Br 2 (g) —> 2NaBr(s)

Answer 1

First, determine the limiting reactant by dividing each mass by its molar mass: Br₂ wins at 0.228 mol, so most of the 2.27 moles of Na are not consumed.

The stoichiometry is (36.5g Br₂) * (1 mol Br₂ / 159.8g Br₂) * (2 mol NaBr / 1 mol Br₂) * (102.9g NaBr / 1 mol NaBr) = 47.0g NaBr = theoretical yield.

Percent yield = 100 actual/theoretical = 100 (40.5/47.0) = 86.2%.