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aalyn [17]
3 years ago
8

How many chloride ions are in 2.6 moles of CaCl2?

Chemistry
1 answer:
Rama09 [41]3 years ago
8 0

Answer:

31.31× 10²³ number of Cl⁻ are present in 2.6 moles of CaCl₂ .

Explanation:

Given data:

Number of moles of CaCl₂ = 2.6 mol

Number of Cl₂ ions = ?

Solution:

CaCl₂  → Ca²⁺ + 2Cl⁻

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

In one mole of CaCl₂ there are two moles of chloride ions present.

In 2.6 mol:

2.6×2 = 5.2 moles

1 mole Cl⁻ =   6.022 × 10²³ number of Cl⁻ ions

5.2 mol ×  6.022 × 10²³ number of Cl⁻  / 1mol

31.31× 10²³ number of Cl⁻

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2 years ago
If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
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Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

Molar mass of O_2 = 32 g/mole

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

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