Respuesta :
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
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
