a. 4 Al + 3 O2 -> 2 Al2O3
b. To calculate the limiting and excess reactant we need to first do a molar ratio between the two reactants, by our balanced equation we know that the ratio is 4:3, which means for every 4 moles of aluminum we will need 3 moles of oxygen to react, so when we have 1.2 moles of oxygen gas, how many moles of aluminum do we have?
4 Al = 3 O2
x Al = 1.2 O2
x = 1.6 moles of Al
How many grams would 1.6 moles of Al be? To answer that we need its molar mass, which is 26.98 g/mol
26.98 g = 1 mol
x grams = 1.6 moles
x = 43.17 grams of Al
We see we have an excess of Al, 60 grams when we only needed 43.17 grams, therefore the limiting reactant is Oxygen and the excess is Aluminum
b. Oxygen gas
c. Aluminum
d. subtracting 60 grams - 43.17 grams
The left over will be 16.83 grams.
e. For every 3 moles of Oxygen we will have 2 moles of Al2O3, therefore for 1.2 moles of Oxygen we will end up with 0.8 moles of Al2O3. The molar mass of Al2O3 is 101.96 g/mol, but when we have only 0.8 moles, the mass will be:
81.57 grams