+0  
 
+1
97
9
avatar+708 

I am really unsure if anyone can please help.

 Feb 8, 2019
 #1
avatar+98060 
0

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

 

 

cool  cool cool

 Feb 8, 2019
 #2
avatar+708 
+1

Thank you for attempting. I will wait for him this is very important. crying I wish i understood it

jjennylove  Feb 8, 2019
edited by jjennylove  Feb 8, 2019
 #3
avatar+99230 
0

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.

 Feb 8, 2019
 #4
avatar
0

ZZZZZZ

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

 Feb 9, 2019
 #6
avatar+17253 
0

I was taking a nap !   ZZZZZzzzzzZZZZZ    Haha .....now I'm going for a run !   cheekycheeky

ElectricPavlov  Feb 9, 2019
 #7
avatar
0

Cool! laugh

Guest Feb 9, 2019
 #5
avatar+17253 
+3

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

 Feb 9, 2019
 #8
avatar+17253 
+3

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

 Feb 9, 2019
 #9
avatar+708 
0

Thank you so much for your help EP , the explanations have really helped. I appreactiate your help seriously. It has helped a TON! THANK YOUsmiley

jjennylove  Feb 11, 2019

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