From the equation we can see that 3 moles of O2 are produced in the reaction
at STP , one mole of gas occupies 22.4 liters (remember that?....we'll need it shortly)
3 x 22.4 = 67.2 liters for the COMPLETE reaction in the equation......we are only going to get some fraction of this reaction because we are only using 212 gm of gold oxide
Au2O3 has a molar weight of 2( gold) + 3(oxygen) = 2(196.97 ) + 3 (15.999) = 441.937 gm (from periodic table)
we only have 212 gm of this stuff which represents 212/441.937 = .4797 moles of Au2O3
ONE MORE THING Look at the equation given: for every 2 moles of Au2O3 used , you GET 3 moles of 02
so you get 3/2 as much O2 in moles as moles of Au2O3 used .......THIS NEEDS TO BE CLEAR.....OK?
.4797 moles AuO3 x 3/2 = .71955 moles of O2 produced
Now remember earlier that an ideal gas at STP = 22.4 liters ???????
22.4 liters/mole X .71955 moles = 16.118 liters of O2 at STP