4C2H3Br3 +Â Â 11O2
--------------------------->Â Â Â 8CO2Â Â
+Â Â Â Â 6H2P +Â Â Â Â 6Br2
1066.8 g        Â
352
g                                                                           Â
479.4g
98.0
g             Â
62.2
g                                                                          Â
?
1066.8 g C2H3Br3Â Â -----------------------> 352 g
O2
98 .0 g C2H3Br3 -------------------------> 352 x 98 / 1066.8
= 32.33 g O2
we need only 32.33 g O2 . but we have 62.2 g O2 so it is excess
reagent .
so limiting regaent is C2H3Br2.
1066.8 g C2H3Br3 --------------------------> 479.4 g Br2
98 g  C2H3Br3 ------------------------------->
479.4 x 98 / 1066.8 = 44.0 g
mass of Br2 formed = 44.0 g