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Newbie

Joined: 29 Sep 2018
Posts: 2

# 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.

Distinguished Member

Joined: 27 Sep 2017
Posts: 269
Location: Berlin, Germany
 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.

Newbie

Joined: 29 Sep 2018
Posts: 2
 Thank you so much.

Distinguished Member

Joined: 27 Sep 2017
Posts: 269
Location: Berlin, Germany
 You wellcome
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