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iren [92.7K]
3 years ago
12

"how many particles would one formula unit of alcl3 produce when dissolved in solution?"

Chemistry
2 answers:
mote1985 [20]3 years ago
7 0
I think you mean “atoms”
And it’s 4 atoms
True [87]3 years ago
3 0

Answer: 4


Justification:


1) The dissolution of a formula unit of AlCl₃ is given by the chemical equation:


AlCl₃ (s) → Al₃⁺ + 3 Cl⁻


So, 1 unit of AlCl₃ yields 1 partilce of Al₃⁺ and 3 particles of Cl⁻.


2) You get that through a mass balance and the oxidation states (balance of charge).


i) Mass balance: the same number of atoms of each species must appear on the reactant and the product sides.


ii) The oxidation state of Al is 3+ and the oxidation state of Cl is 1-, so the total net charge is 3+ + 3(1-) = 3 - 3 = 0.

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Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
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Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

Molar mass of KI = 166 g/mole

\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

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\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

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