Overall balanced reaction and limiting reagent
Post new topic Reply to topic
   Chemical Community Forum Index : Chemistry forum
View previous topic :: View next topic  
Author Message
JamesVonBurg
Newbie
Newbie


Joined: 29 Sep 2018
Posts: 2
Reply with quoteFind all posts by JamesVonBurg

Overall balanced reaction and limiting reagent


Hi,

I was asked to create the overall balanced equation for....

2Al(s) + 2koh(aq) + 6H2O(l) = 2K[Al(OH)4](aq) + 3H2(g)
2K[Al(OH)4](aq) + H2SO4(aq) = 2AL(OH)3(s) + K2SO4(aq) + 2H2)(l)
2Al(OH)3(s) + 3H2SO4(aq) = 2Al^3+(aq) + 3SO4^2-(aq) + 6H2)(l)
K^+(aq) + Al^3+(aq) + 2SO4^2-(aq) + 12H2O(l) = KAl(SO4)2 x 12H2O(s)

And I got,

2Al(s) + 2KOH(aq) + 4H2SO4(aq) + K(aq) + 10H2O(l) = 3H2(g) + K2SO4(aq) + Al^3+(aq) + SO4^2-(aq) + KAl(SO4)2 x 12H2O(s)

I was then asked to find the %yield so I figured I needed to find the limiting reagent so stoichiometry, but I am confused on how to balance this gigantic equation in order to find the limiting reagent.

Please provide a solution for this as I am totally lost in this problem

Thanks and sorry for the gigantic equation.
Back to top
View user's profileSend private message
ChenBeier
Sr. Staff Member
Sr. Staff Member


Joined: 27 Sep 2017
Posts: 226
Location: Berlin, Germany
Reply with quoteFind all posts by ChenBeier


The final equation is

2 Al + 2 KOH + 4 H2SO4 + 22 H2O => 2( KAl(SO4)2 * 12 H2O) + 3 H2

Now it depends how much you have from each edukt, then you can calculate the yield and the limiting componente.
Back to top
View user's profileSend private message
JamesVonBurg
Newbie
Newbie


Joined: 29 Sep 2018
Posts: 2
Reply with quoteFind all posts by JamesVonBurg


Thank you so much.
Back to top
View user's profileSend private message
ChenBeier
Sr. Staff Member
Sr. Staff Member


Joined: 27 Sep 2017
Posts: 226
Location: Berlin, Germany
Reply with quoteFind all posts by ChenBeier


You wellcome
Back to top
View user's profileSend private message
Display posts from previous:   
Post new topic Reply to topic
   Chemical Community Forum Index : Chemistry forum Page 1 of 1

 
Jump to:  

Chemical portal