Question

how many moles of potassium hydroxide are needed to completely react with 1.73 miles of aluminum sulfate according to the following equation al2(so4)3+6koh=2al(oh)3+3k2so4

Answer 1

Answer;

=10.38  moles KOH  

Explanation;

The balanced equation.  

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

From the equation;

1 mole of aluminum sulfate requires 6 moles of potassium hydroxide.  

Moles of Aluminium sulfate;  1.73 moles

Moles of KOH;

1 mol Al2(SO4)3 : 6 mol KOH = 1.73  mol Al2(SO4)3 : x mol KOH  

Thus; x =  (6 × 1.73)

              =10.38 moles KOH

Answer 2

Hello!

The answer is: 10.39 moles of potassium hydroxide (KOH) are needed to completely react with 1.73 moles of aluminum sulfate.

Why?

Since the equation is already balanced, we can calculate how many moles of potassium hydroxide are needed.

We must remember that in every chemical reaction, the mass is conserved, it means that the total mass of the reactants is equal to the total mass of the products.

So, from the equation, we can know that 1 mole of aluminum sulfate reacts with 6 moles of potassium, so if we need to know how many moles of hydroxide are needed to react with 1.73 moles of aluminum sulfate, we can write the following relation:

$\frac{1 mol Al2(SO4)3}{6 mol KOH}=\frac{1.73 mol Al2(SO4)3}{n(KOH)} \\\\n(KOH)=\frac{6 mol KOH*1.73molAl2(SO4)3}{1 mol Al2(SO4)3}\\\\n(KOH)=10.386molKOH$

So, 10.39 moles of potassium hydroxide (KOH) are needed to completely react with 1.73 moles of aluminum sulfate.

Have a nice day!