Consider the reaction:

Question

Consider the reaction 

i. Which is the limiting reactant if we start with 114 g of Al (26.98 g/mol) and 186 g of (70.9 g/mol)?

ii. How much  (133.34 g/mol) can be made at most from these amounts of reactants?

iii. Which reactant is in excess?

iv. How many grams of the excess reactant are left over?

Choose the correct set of answers to questions i-iv:

Answer 1

2Al + 3Cl₂ ==> 2AlCl₃ ... balanced equation

moles Al = 114 g x 1 mol/26.98 g = 4.225 moles Al

moles Cl₂ = 186 g x 1 mol/70.9 g = 2.623 moles Cl₂

i. Since it takes 3 moles of Cl2 for 2 moles Al, the Cl₂ is in limiting supply

ii. 2.623 mol Cl₂ x 2 mol AlCl₃ / 3 mol Cl₂ x 133.34 g AlCl₃/mol = 233 g AlCl₃ produced

iii. Since Cl2 is limiting, Al is in excess

iv. moles of Al reacted = 2.623 mol Cl₂ x 2 mol Al / 3 mol Cl₂ = 1.75 mol Al reacted

mass of Al reacted = 1.75 mol x 26.98 g/mol = 47.2 g Al reacted

mass of Al left over = 114 g - 47.2 g = 66.8 g = 67 g of Al left over