Chemistry Equilibrium Question?

Chemistry
Tutor: None Selected Time limit: 1 Day

Hello. So in this chem. equation both sides have 4KO2(s) + 2H2O(g) 4KOH(s) + 3O2(g) 1 s and 1 g.. so how do I balance it? I knoe K = [C]c[D]d divided by [A]a[B]b but this..?

Dec 11th, 2014

Since your equation is balanced, you need to know one thing: the # in front of each molecule, called the coefficient, is the number of MOLES of that molecule you need for the balanced equation to work. In other words you are reacting 4 moles of KO2 and 2 moles of water to make 3 moles of diatomic oxygen and 4 moles of KOH. 

If you were to make 63.92 moles of KOH, then divide 63.92 by 4 (because that's how many you have in the balanced rxn) (15.98) and multiply all the other coefficients by that to remain balanced. 
Now you have.... 
63.92 KO2(s) + 31.96 H2O(l) ---> 47.94 O2(g) + 63.92 KOH(s) 
Look at the coefficient for KO2 - now its the same as KOH, 62.92 moles! That's your answer. 

As to the limiting reactant question, look at your balanced equation. You have a 2:1 ratio in the reactants (4 KO2: 2 H2O) and a nearly equal ratio (3:4) in products. 

10.94 into 34.98 = 3.2 (3.197) so isn't an even number. 
We'll have to plug in the numbers and balance the equation the old fashioned way! 
34.98 KO2 + 10.94 H2O => 22.96 O2 + 34.98 KOH 
K 34.98 both sides, O 80.9 both sides; H 21.88 on left, 34.98 right 

I can't get the Hydrogen to balance with these numbers in the reactant column, so hydrogen (water) is definitely the limiting reactant. 

34.98 KO2 + 17.49 H2O => 26.235 O2 + 34.98 KOH balanced.

Dec 11th, 2014

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