Answer
See answer below.
Work Step by Step
a) From the graph: mass of product: $11\ g$
Mass of $Br_2$: $11-2=9\ g$
b) Number of moles of Fe: $2.0\ g\div 55.845\ g/mol=0.0358\ mol$
Number of moles of bromine: $9.0\ g\div 159.81\ g/mol=0.0563\ mol$
Ratio: $1.57\approx 3/2$
c) From item b: $FeBr_3$
d)$2\ Fe+ 3\ Br_2\rightarrow 2\ FeBr_3$
e) Iron (III) bromide.
f) I - correct
II - false, iron is in excess.
III - false, iron is in excess.
IV - false, that's not how percent yield is calculated.
