If 5 moles of aluminum Al was reacted with 10 moles bromine Br2, all five moles of aluminum would react, with 7.5 moles bromine. (2:3 mole ratio)

Now assume 3 moles Al and 4 moles Br2 react

a) which chemical is the limiting reactant?

b) what chemical must be the excess reactant

c) How much (in moles) AlBr3 gets produced?

d) If all the limiting reactant gets used up, how much of the excess reactant is left?

Q1)

we have been told that molar ratio of Al to Br₂ is in the 2:3 ratio

the balanced equation for the reaction between Al and Br₂ is as follows;

2Al + 3Br₂ - - > 2AlBr₃

therefore the molar ratio of Al to Br₂ is 2:3 as stated previously

when 3 mol of Al react with 4 mol of Br₂

we have to find the limiting reactant. Limiting reactant is the reagent that is fully consumed during the reaction.

If we assumed Al to be the limiting reactant

if 2 mol of Al react with 3 mol of Br₂

then 3 mol of Al would react with - 3/2 x 3 = 4.5 mol of Br₂

however only 4 mol of Br₃ is present, this means that Br₂ is the limiting reactant.

Answer for limiting reactant is Br₂

Q2)

excess reactant is the reagent in which only a fraction of the amount present is used up in the reaction

when Br₂ is the limiting reactant

if 3 mol of Br₂ react with 2 mol of Al

and 4 mol of Br₂ present

then 4 mol of Br₂ react with - 2/3 x 4 = 2.67 mol of Al

but 3 mol of Al is present, more than the required amount.

Therefore Al is the excess reactant

Q3)

according to the balanced equation mentioned previously,

stoichiometry of Br₂ to AlBr₃ is 3:2

the amount of product formed depends on amount of limiting reactant present.

when 3 mol of Br₂ reacts - then 2 mol of AlBr₃ is formed

now since 4 mol of the limiting reactant is present

then when 4 mol of Br₂ reacts - 2/3x 4 = 2.67 mol of AlBr₃ is formed

number of AlBr₃ moles formed are - 2.67 mol

Q4)

limiting reactant the whole amount is consumed.

when Br₂ is the limiting reactant, 4 moles of Br₂ present are fully used up.

as calculated above, 4 mol of Br₂ reacts with 2.67 mol of Al

and 3 moles of Al provided.

Used amount of Al - 2.67 mol

Therefore amount of Al in excess = 3 - 2.67 = 0.33 mol

amount of excess reactant left over after the reaction is 0.33 mol