Limiting Reactions

Question

Use the following equation for the oxidation of aluminum in the following problems.

4 Al + 3 O₂ ---> 2 Al₂O₃

_{a. Which reactant is limiting if 0.32 mol of Al and 0.26 mol of O2 are available?}

_{b. How many moles of Al2O3 are formed from the reaction of 6.38 x 10}^-3 _{mol of O2 and 9.15 10}^-3 _{mol of Al?}

_{c. If 3.17 g of Al and 2.55 g of O2 are available, which reactant is limiting?}

Answer 1

a) (0.32 mol Al) * (2 mol Al₂O₃ /4 mol Al) = 0.160 mol Al₂O₃. (0.26 mol O₂) * (2 mol Al₂O₃ /3 mol O₂) = 0.173 mol Al₂O₃. Al metal is limiting because there is not enough Al to consume all the oxygen gas.

b) (6.38*10^-3 mol O₂) * (2 mol Al₂O₃ /3 mol O₂) = 4.25*10^-3 mol Al₂O₃. (9.15*10^-3 mol Al) * (2 mol Al₂O₃ /4 mol Al) = 4.58*10^-3 mol Al₂O₃. Oxygen gas is limiting, producing 4.25*10^-3 mol Al₂O₃.

c) (3.17g Al) *(1 mol Al /26.98g Al) * (2 mol Al₂O₃ /4 mol Al) = 0.0587 mol Al₂O₃.

(2.55g O₂) * (1mol O₂ /32.0g O₂) * (2 mol Al₂O₃ /3 mol O₂) = 0.0531 mol Al₂O₃. Oxygen gas is limiting.