Answer
59.1 g
Work Step by Step
Grams of sodium iodide to be used=
$50.0\,g\times\frac{1\,mol\,I_{2}}{253.8\,g}\times\frac{2\,mol\,NaI}{1\,mol\,I_{2}}\times\frac{149.89\,g}{1\,mol\,NaI}$
$=59.1\,g$
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