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e-lub [12.9K]
3 years ago
9

How many liters of fluorine gas are needed to form 879 L of sulfur hexafluoride gas if the following reaction takes place at 2.0

0 atm and 273.15 K
Chemistry
1 answer:
Ghella [55]3 years ago
4 0

Answer:

2637.3 litters of fluorine gas are needed to produce 879 litters of sulfur hexafluoride.

Explanation:

Given data:

Volume of SF₆ = 879 L

Pressure = 2 atm

Temperature = 273.15 k

Volume of fluorine required = ?

Solution:

Balance chemical equation:

    S (s) + F₂(g)    →  SF₆(g)

First of all we will calculate the moles of SF₆.

  PV = nRT

  n = PV/RT

  n = 2. atm× 879L / 0.0821 L. atm. mol⁻¹. K⁻¹  × 273.15 K

  n = 1758 atm. L/ 22.43 L. atm. mol⁻¹

  n = 78.4 mol

78.4 moles of SF₆ will produce.

Now we will compare the moles of SF₆ and fluorine from balance chemical equation.

      SF₆   :   F

        1     :    3

      78.4  :   3/1 × 78.4 = 235.2 moles

Now we will calculate the volume of fluorine.

   PV = nRT

  V  = nRT / P

  V= 235.2 mol. 0.0821 L. atm. mol⁻¹. K⁻¹  × 273.15 K / 2 atm

  V = 5274. 5 / 2

  V = 2637.3 L

2637.3 litters of fluorine gas are needed to produce 879 litters of sulfur hexafluoride.

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What pressure will be produced when 2.0 moles of N2 gas is heated to 68oC in a container that holds 1.25 of gas?
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The pressure of the nitrogen gas produced is determined as 44.77 atm.

<h3>What is the pressure of the Nitrogen gas?</h3>

The pressure of the nitrogen gas is determined from ideal gas equation, as shown below;

PV = nRT

P = nRT/V

where;

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  • R is ideal gas constant = 0.08205 L.atm/mol.K
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P = (2 x 0.08205 x 341)/(1.25)

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Learn more about pressure here: brainly.com/question/25736513

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Niven wants to calculate the mass of mgo that is produced by burning of 28.0g of mg. What is the first step in nivens calculatio
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Answer: -

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3 years ago
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