A reaction of 41.1 g of Na and 59.7 g of Br, yields 73.9 g of...

Question

Question

A reaction of 41.1 g of Na and 59.7 g of Br, yields 73.9 g of NaBr. What is the percent yield? 2 Na(s) + Br2(g) → 2 NaBr(s)

Answer 1

Answer #1

Solution-
The reaction can be written as

2Na + Br2 ----> 2NaBr

Now the moles of Na = 41.1/23

= 1.78

And the moles of Br2 = 59.7/159.8

= 0.373

From the reaction it can be said that one mole of Br2 reacts with two moles of Na

Therefore here Br2 is limiting reagent.

The moles of NaBr formed = 2 * moles of Br2 = 0.746 moles

And the mass of NaBr formed = moles * MOl weight = 0.746*103

= 76.838 grams

So the percent yield = (Actual yield/Theoretical yiwle)*100

= (73.9/76.838 )*100

= 96.17 %

Answer 2

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