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worty [1.4K]
3 years ago
10

how many moles of potassium hydroxide are needed to completely react with 1.73 miles of aluminum sulfate according to the follow

ing equation al2(so4)3+6koh=2al(oh)3+3k2so4
Chemistry
2 answers:
kifflom [539]3 years ago
8 0
<h2>Hello!</h2>

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

<h2>Why?</h2>

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!

Masteriza [31]3 years ago
4 0
<h3><u>Answer; </u></h3>

=10.38  moles KOH  

<h3><u>Explanation</u>;</h3>

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)

              <u> =10.38 moles KOH </u>

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Consider the reaction given below.
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  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

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This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

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When 1. 0 l of 0. 00010 m naoh and 1. 0 l of 0. 0014 m mgso4 are mixed, would a precipitate be formed? show work
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<h3>What is a precipitate?</h3>

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First, we determine the concentration of magnesium and hydroxide

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