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vlabodo [156]
3 years ago
14

What volume is occupied by 8.7 g of chlorine gas, Cl2, at 23°C and 1.15 atm pressure

Chemistry
1 answer:
Harrizon [31]3 years ago
4 0

Answer:

V = 5.17L

Explanation:

Mass of gas = 8.7g

T = 23°C = (23 + 273.15)K = 296.15K

P = 1.15 atm

V = ?

R = 0.082atm.L / mol.K

From ideal gas equation

PV = nRT

P = pressure of the gas

V = volume of the gas

n = no. Of moles

R = ideal gas constant

T = temperature of the gas

no of moles = mass / molar mass

Molar mass of Chlorine = 35.5g / mol

No. Of moles = 8.7 / 35.5

No. Of moles = 0.245 moles

PV = nRT

V = nRT / P

V = (0.245 * 0.082 * 296.15) / 1.15

V = 5.9496 / 1.15

V = 5.17L

The volume of the gas is 5.17L

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Number of moles:

The number of moles is the product of molarity of the solution and its volume. The formula is expressed as:

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Calculations:

Step 1:

The molecular formula of silver nitrate is AgNO3. The number of moles of silver nitrate is calculated as:

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The molecular formula potassium chromate is K2CrO4. The number of moles of potassium chromate is calculated as:

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The balanced chemical reaction between AgNO3 and K2CrO4 is:

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The required number of moles of K2CrO4 = 0.075 mol/2 = 0.0375 mol

The given number of moles of K2CrO4 (0.04 mol) is more than the required number of moles (0.0375 mol). Therefore, AgNO3 is the limiting reagent.

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According to the reaction, the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Hence, the number of moles of Ag2CrO4 formed is 0.0375 mol.

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The mass of Ag2CrO4 is calculated as:

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