What is the percent yield for the reaction between 45.9 g of #NaBr# and excess chlorine gas to produce 12.8 g of #NaCl# and an unknown quantity of bromine gas?

Question

7804 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-07T09:52:21

Approx. 50%...........

We interrogate the given reaction.....

#NaBr(aq) + 1/2Cl_2(aq)rarrNaCl(aq)+1/2Br_2(aq)#

Bromide ion is oxidized, and chlorine is reduced........

#"Moles of NaBr"=(45.9*g)/(102.89*g*mol^-1)=0.446*mol#

#"Moles of NaCl"=(12.8*g)/(58.44*g*mol^-1)=0.219*mol#

And the yield is simply the quotient of these molar quantities.....

#"% Yield"="Moles of NaCl"/"Moles of NaBr"xx100%#

#=(0.219*mol)/(0.446*mol)xx100%=??%#