A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield? 2Na(s)+Br2(g)⟶2NaBr(s)

The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

Now we have to calculate the moles of

The balanced chemical reaction is,

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = moles of Sodium bromide

Now we have to calculate the percent yield of reaction

Therefore, the percent yield is, 93.4%

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