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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)

User Joe Doyle
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1 Answer

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Answer: The percent yield is, 93.4%

Step-by-step explanation:

First we have to calculate the moles of Na.


\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=(41.9g)/(23g/mole)=1.82moles

Now we have to calculate the moles of
Br_2


{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =(30.3g)/(160g/mole)=0.189moles


{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =(36.4g)/(103g/mole)=0.353moles

The balanced chemical reaction is,


2Na(s)+Br_2(g)\rightarrow 2NaBr

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

So, 0.189 moles of bromine react with =
(2)/(1)* 0.189=0.378 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 =
(2)/(1)* 0.189=0.378 moles of Sodium bromide

Now we have to calculate the percent yield of reaction


\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}* 100=(0.353 mol)/(0.378)* 100=93.4\%

Therefore, the percent yield is, 93.4%

User Parapluie
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