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How many moles of sodium bromide can be produced from the reaction of 1. 03 moles of bromine gas with 0. 650 moles of sodium? 2 na(s) br2(g) ® 2 nabr(s)

Sagot :

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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