If anyone would be able to help

I can't answer these, jenny....but....I believe that EP knows some Chemistry.....he will probably be on in a while.....

Hi Jenny, I cannot help either but I would like to suggest that you use a title like:

Chemistry - help needed please.

It is better when people use less generic titles.

ZZZZZZ

It’s about time EP! Jenny has been waiting for hours and hours!

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

Ooops....didn'y answer all of the questions...

#2    2 moles of  AuO3  produce 4 moles of Au

      .4797 moles Au2O3  x 2 = .9594 moles Au produced

#3  85 grams of Au = 85/196.96 mole          1/2 as many moles Au2O3 is needed

      85/196.96   x  1/2 = .4315 moles

we already found Au2O3 moles weigh 441.937 gm

     .4315 mole x 441.937 gm/mole = 190.72 gms of Au2O3 are needed