# How many grams of Al2O₃ were decomposed to produce the 0.455 L of O₂ at 22°C and 724mm Hg?

To solve this problem, use the ideal gas law to determine the mole of oxygen produced (= 0.0179 moles). Use stoichiometry to determine the moles of aluminum oxide decomposed (= 0.012 moles). Using the molar mass (102 g/mol), convert the moles of aluminum oxide to g (= 1.224 g).

The balanced chemical equation for the decomposition of aluminum oxide (Al2O3) can be written as

2Al2O3 -> 4Al + 3O2.

According to this equation, 2 moles of aluminum oxide decomposes to form 4 moles of aluminum and 3 moles of oxygen gas.

Here, we are given information about the product oxygen. 0.455 L of oxygen are produced at 22 degrees Celsius and 724 mm Hg.

We need to convert it to moles of oxygen produced first, and then using the stoichiometry, we can figure out the moles of aluminum oxide consumed.

Using the ideal gas law, PV = nRT,

or n = PV/RT,

where, P = pressure of oxygen = 724 mm Hg = 724/760 atm = 0.953 atm,

V = volume of oxygen = 0.455 L,

R = universal gas constant = 0.0821 L/atm/mol/K,

and T = temperature in absolute scale = 22 degrees Celsius = 22 + 273 K = 295 K.

Thus, moles of oxygen produced, n = PV/RT = (0.953 * 0.455) / (0.0821 * 295) = 0.0179 moles oxygen.

Using stoichiometry, 3 moles of oxygen are generated when 2 moles of aluminum oxide gets decomposed, or, 1 mole of oxygen is generated for 2/3 moles of aluminum oxide decomposed.

Thus, 0.0179 moles of oxygen are produced for 2/3 * 0.0179 moles of aluminum oxide decomposed, which is equal to 0.012 moles.

Thus, 0.012 moles of aluminum oxide is decomposed to produce the given volume of oxygen at the given pressure and temperature.

The molecular mass of aluminum oxide is 102 g/mol.

Therefore, 0.012 x 102 g = 1.224 g of aluminum oxide are decomposed for the given product.

Hope this helps.