Answer : The mass of
is, 295.323 grams
Solution :
First we have to calculate the moles of
.

Now we have to calculate the moles of NaBr.
The balanced chemical reaction is,

From the balanced reaction we conclude that
As, 2 moles of
react with 2 moles of 
So, 2.87 moles of
react with 2.87 moles of 
Now we have to calculate the mass of NaBr.


Therefore, the mass mass of
is, 295.323 grams