Question

What are the missing coefficients for the skeleton equation below? Al2(SO4)3(aq) + KOH(aq) ® Al(OH)3(aq) + K2SO4(aq)

Answer 1

You have to balance the chemical equation:


Al2(SO4)3(aq) + KOH(aq) ---> Al(OH)3(aq) + K2SO4(aq)


To balance that equation you can start by balancing the SO4 radicals. That requires that you place a 3 coefficient before K2SO4 (in the right side):


Al2(SO4)3(aq) + KOH(aq) ---> Al(OH)3(aq) + 3K2SO4(aq)


Now write a 6 in front of KOH in the left side, to balance K


 
Al2(SO4)3(aq) + 6KOH(aq) ---> Al(OH)3(aq) + 3K2SO4(aq)



Now wirte a 2 before Al(OH)3 in the right side to balance the OH radicals


Al2(SO4)3(aq) + 6KOH(aq) ---> 2Al(OH)3(aq) + 3K2SO4(aq) 


Now you can verifiy that the equation is balanced.


The answer is: Al2(SO4)3(aq) + 6KOH(aq) ---> 2Al(OH)3(aq) + 3K2SO4(aq)

Answer 2

The skeleton equation is an equalisation that represents all chemicals including the reactants and the products in a formula manner.

In the question reaction between aluminium sulphate and potassium hydroxide is given as:

Al₂(SO₄)₃(aq) + KOH(aq) ---> Al(OH)₃(aq) + K₂SO₄(aq)

The given equation will be balanced as:

1. By balancing the SO₄. Add 3 as a coefficient on the right side of the equation before K₂SO₄.

Al₂(SO₄)₃(aq) + KOH(aq) ---> Al(OH)₃(aq) + 3K₂SO₄(aq)

2. Now balance potassium(K) by adding 6 as a coefficient in front of KOH on the left of the equation.

Al₂(SO₄)₃(aq) + 6KOH(aq) ---> Al(OH)₃(aq) + 3K₂SO₄(aq)

3. Balance OH by adding 2 as a coefficient in front of Al(OH)₃ on the right side of the equation.

Al₂(SO₄)₃(aq) + 6KOH(aq) ---> 2Al(OH)₃(aq) + 3K₂SO₄(aq)

Therefore, the balanced equation is:

Al₂(SO₄)₃(aq) + 6KOH(aq) ---> 2Al(OH)₃(aq) + 3K₂SO₄(aq)

To learn more about balancing the equation follow the given link:

https://brainly.com/question/12405075