Consider the reaction: 2 al + 3br2 → 2 albr3 suppose a reaction vessel initially contains 5.0 mole al and 6.0 mole br2. what is in the reaction vessel once the reaction has occurred to the fullest extent possible?

The balanced equation for the above reaction is as follows;
2Al + 3Br₂ --> 2AlBr₃
stoichiometry of Al to Br₂ is 2:3
there are 5.0 mol of Al and 6.0 mol of Br₂
if Al is the limiting reactant,
if 2 mol of Al reacts with 3 mol of Br₂
then 5.0 mol of Al reacts with - 3/2 x 5 = 7.5 mol of Br₂
but only 6 mol of Br₂ is present therefore Br₂ is the limiting reactant.
if 3 mol of Br₂ reacts with 2 mol of Al 
then 6.0 mol of Br₂ reacts with - 2/3 x 6 = 4 mol of Al

the molar ratio of Al to AlBr₃ is 2:2
the number of moles of AlBr₃ formed is 4 mol of AlBr₃
amount of excess Al - 5.0 mol - 4.0 mol = 1 mol 
therefore at the end of the reaction, 

1 mol of Al and 4 mol of AlBr₃ in the reaction vessel

The answer is c fish

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