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coldgirl [10]
3 years ago
9

At 25 °c only 0.0260 mol of the generic salt ab3 is soluble in 1.00 l of water. what is the ksp of the salt at 25 °c?

Chemistry
2 answers:
Schach [20]3 years ago
6 0

<u>Answer:</u> The K_{sp} of the given salt at 25°C is 1.23\times 10^{-5}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of AB_3 = 0.0260 mol

Volume of solution = 1.00 L

Putting values in above equation, we get:

\text{Molarity of }AB_3=\frac{0.0260mol}{1L}=0.0260M

The chemical equation for the ionization of AB_3 salt follows:

AB_3(aq.)\rightarrow A^{3+}(aq.)+3B^-(aq.)

The expression of K_{sp} for above reaction is written as:

K_{sp}=[A^{3+}][3B^{-}]^3

We are given:

[A^{3+}]=0.0260M

[B^-]=(3\times 0.0260)=0.078M

Putting values in above expression, we get:

K_{sp}=(0.0260)\times (0.078)^3\\\\K_{sp}=1.23\times 10^{-5}

Hence, the K_{sp} of the given salt at 25°C is 1.23\times 10^{-5}

Sladkaya [172]3 years ago
4 0
Given:

At 25 degrees Celcius:

amount of generic salt AB3 = 0.0260 moles
Volume of solvent = 1.0 L water

Set up a balanced chemical equation:

AB3 =========> A3+ + 3B-
0.0260M             0.0260   0.078

Ksp = [A][B]^3
Ksp = 1.23 x 10^-5<span />
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Taking into account the reaction stoichiometry, you can observe that:

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  • the mole ratio between phosphine and calcium phosphide is 2 mol PH₃ over 1 mol Ca₃P₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Ca₃P₂ + 6 H₂O  → 3 Ca(OH)₂ + 2 PH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Ca₃P₂:1 mole
  • H₂O: 6 moles
  • Ca(OH)₂: 3 moles
  • PH₃: 2 moles

The molar mass of the compounds is:

  • Ca₃P₂: 182 g/mole
  • H₂O: 18 g/mole
  • Ca(OH)₂: 74 g/mole
  • PH₃: 34 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Ca₃P₂: 1 mole ×182 g/mole= 182 grams
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  • PH₃: 2 moles ×34 g/mole= 68 grams

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Then, by reaction stoichiometry, you can observe that:

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Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g)  →  3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

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