the answer is in the picture, btw the molar mass for the first one is wrong, it should be 77.98, and the final product is 2.32
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First the theoretical yield of Nabr
by use of mole ratio between FeBr3 and NaBr which is 2:6 the theoretical yield
=2.36 x6/2= 7.08 moles
the % yield = actual yield/ theoretical yield x 100
that is 6.14/7.08 x100= 86.72%
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