I am really unsure if anyone can please 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.
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