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.