There are 0. 650 moles of sodium bromide can be produced from the reaction of 1. 03 moles of bromine gas with 0. 650 moles of sodium .
Calculation ,
To find the number of moles of sodium bromide we have to write balanced chemical equation first .
[tex]2Na (s)+ Br_{2}(g)[/tex] → 2NaBr (s)
from the equation we can conclude that 2 moles sodium react with one mole of bromine to form 2 mole of sodium bromide .
But 1. 03 moles of bromine gas ( that is equal to the number of moles required in the equation) with 0. 650 moles of sodium ( that is half of required mole ) is given .
So, sodium is present as limiting reactant .
So , the number of sodium bromide formed is equal to the number of moles of sodium atom.
Hence , the number of sodium bromide formed = 0. 650 moles
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