13.5 grams of aluminum reacts with 31.0 grams of oxygen

What mass of aluminum oxide forms assuming oxygen is the limiting reactant?

Group Activity 2 Activity 1: Aluminum cookware is often referred to as "anodized aluminum. The anodizing process puts a layer of aluminum oxide on the aluminum, protecting it from corrosion. Consider the reaction of 12.50 grams of aluminum with 12.50 grams of oxygen to form aluminum oxide How many grams of product are formed? What is the stated goal of the problem? a. b. What information is available and what information is needed? What is the formula for aluminum oxide? Write and balance the chemical equation for the process. c. d. Calculate the number of moles of aluminum in 12.50g Al. e. Calculate the number of moles of oxygen in 12.50g O2 f. How many moles of product would be expected if oxygen is the limiting reactant? g. How many moles of product would be expected if aluminum is the limiting reactant? h. Which is the limiting reactant? i. Determine the moles of product formed. J Determine the number of grams of product formed.

3- Powdered aluminum and iron(II) oxide react with a large evolution of heat according to the following equation 2 Al + Fe,03 → Alo, + 2 Fe In one experiment, 0.360 mole of Al was added to 0.200 mole of Fe2O, a) Which reactant, if either, was the limiting reactant? b) Calculate the theoretical yield (in moles) of iron. 4- The following reaction occurs when aluminum is heated with oxygen: A reacting mixture contains 50.0 g aluminum and 50.0 g of oxygen. a) Find the limiting reagent. b) How many grams of aluminum oxide will form? c) How many grams of excess reagent will remain?